Does the series $sumlimits_n=1^inftyfrac (-1)^n log(n)sqrtn$ converge?
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up vote
2
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The following series converge or diverge?
$$sum_n=1^inftyfrac (-1)^n log(n)sqrtn$$
I have proved that this series diverges absolutely.
I tried to use Leibniz criterion:
- $a_n >0$ definitively.
- The limit of $a_n=0$ (as n tends to infinity).
- $log(n)/sqrt n >log(n+1)/sqrtn+1$ definitively
it's ok?
real-analysis sequences-and-series
edited Aug 30 at 5:49
Robson
47320
asked Jan 23 '13 at 14:20
Agenog
608722
add a comment |Â
up vote
2
down vote
favorite
The following series converge or diverge?
$$sum_n=1^inftyfrac (-1)^n log(n)sqrtn$$
I have proved that this series diverges absolutely.
I tried to use Leibniz criterion:
- $a_n >0$ definitively.
- The limit of $a_n=0$ (as n tends to infinity).
- $log(n)/sqrt n >log(n+1)/sqrtn+1$ definitively
it's ok?
real-analysis sequences-and-series
edited Aug 30 at 5:49
Robson
47320
asked Jan 23 '13 at 14:20
Agenog
608722
That is what you need to show, yes. Best to be explicit about what $a_n$ is from the start. You obviously can't just assert (2) and (3), but have to prove them.
– Thomas Andrews
Jan 23 '13 at 14:28
1) and 2) are obvious; 3) it's proved with some passages and with passage to the limit. But wolfram said: "sum does not converge, ratio test inconclusive, root test inconclusive"
– Agenog
Jan 23 '13 at 14:33
1
How can you prove (3) by passing to the limit, when you are trying to prove it for specific values, or do you just mean the function is strictly dereasing as a continuous function?
– Thomas Andrews
Jan 23 '13 at 14:36
you can arrive to $log^2(n)> n*log^2(1+1/n) and with passage to limit the first part --> +inf, the second part of inequality -->0 so definitively I have proved the inequality
– Agenog
Jan 23 '13 at 14:58
1
Again, you say "passage to the limit." Part of it is that you misformatted that comment, but I also just don't understand the thrust of your argument. What limit? If you send $ntoinfty$ how are you showing that $log(n)/sqrt n >log(n+1)/sqrtn+1$?
– Thomas Andrews
Jan 23 '13 at 15:18
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The following series converge or diverge?
$$sum_n=1^inftyfrac (-1)^n log(n)sqrtn$$
I have proved that this series diverges absolutely.
I tried to use Leibniz criterion:
- $a_n >0$ definitively.
- The limit of $a_n=0$ (as n tends to infinity).
- $log(n)/sqrt n >log(n+1)/sqrtn+1$ definitively
it's ok?
real-analysis sequences-and-series
edited Aug 30 at 5:49
Robson
47320
asked Jan 23 '13 at 14:20
Agenog
608722
The following series converge or diverge?
$$sum_n=1^inftyfrac (-1)^n log(n)sqrtn$$
I have proved that this series diverges absolutely.
I tried to use Leibniz criterion:
- $a_n >0$ definitively.
- The limit of $a_n=0$ (as n tends to infinity).
- $log(n)/sqrt n >log(n+1)/sqrtn+1$ definitively
it's ok?
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Aug 30 at 5:49
Robson
47320
asked Jan 23 '13 at 14:20
Agenog
608722
edited Aug 30 at 5:49
Robson
47320
asked Jan 23 '13 at 14:20
Agenog
608722
edited Aug 30 at 5:49
Robson
47320
edited Aug 30 at 5:49
Robson
47320
edited Aug 30 at 5:49
Robson
47320
47320
asked Jan 23 '13 at 14:20
Agenog
608722
asked Jan 23 '13 at 14:20
Agenog
608722
asked Jan 23 '13 at 14:20
Agenog
608722
608722
That is what you need to show, yes. Best to be explicit about what $a_n$ is from the start. You obviously can't just assert (2) and (3), but have to prove them.
– Thomas Andrews
Jan 23 '13 at 14:28
1) and 2) are obvious; 3) it's proved with some passages and with passage to the limit. But wolfram said: "sum does not converge, ratio test inconclusive, root test inconclusive"
– Agenog
Jan 23 '13 at 14:33
1
How can you prove (3) by passing to the limit, when you are trying to prove it for specific values, or do you just mean the function is strictly dereasing as a continuous function?
– Thomas Andrews
Jan 23 '13 at 14:36
you can arrive to $log^2(n)> n*log^2(1+1/n) and with passage to limit the first part --> +inf, the second part of inequality -->0 so definitively I have proved the inequality
– Agenog
Jan 23 '13 at 14:58
1
Again, you say "passage to the limit." Part of it is that you misformatted that comment, but I also just don't understand the thrust of your argument. What limit? If you send $ntoinfty$ how are you showing that $log(n)/sqrt n >log(n+1)/sqrtn+1$?
– Thomas Andrews
Jan 23 '13 at 15:18
add a comment |Â
That is what you need to show, yes. Best to be explicit about what $a_n$ is from the start. You obviously can't just assert (2) and (3), but have to prove them.
– Thomas Andrews
Jan 23 '13 at 14:28
1) and 2) are obvious; 3) it's proved with some passages and with passage to the limit. But wolfram said: "sum does not converge, ratio test inconclusive, root test inconclusive"
– Agenog
Jan 23 '13 at 14:33
1
How can you prove (3) by passing to the limit, when you are trying to prove it for specific values, or do you just mean the function is strictly dereasing as a continuous function?
– Thomas Andrews
Jan 23 '13 at 14:36
you can arrive to $log^2(n)> n*log^2(1+1/n) and with passage to limit the first part --> +inf, the second part of inequality -->0 so definitively I have proved the inequality
– Agenog
Jan 23 '13 at 14:58
1
Again, you say "passage to the limit." Part of it is that you misformatted that comment, but I also just don't understand the thrust of your argument. What limit? If you send $ntoinfty$ how are you showing that $log(n)/sqrt n >log(n+1)/sqrtn+1$?
– Thomas Andrews
Jan 23 '13 at 15:18
That is what you need to show, yes. Best to be explicit about what $a_n$ is from the start. You obviously can't just assert (2) and (3), but have to prove them.
– Thomas Andrews
Jan 23 '13 at 14:28
That is what you need to show, yes. Best to be explicit about what $a_n$ is from the start. You obviously can't just assert (2) and (3), but have to prove them.
– Thomas Andrews
Jan 23 '13 at 14:28
1) and 2) are obvious; 3) it's proved with some passages and with passage to the limit. But wolfram said: "sum does not converge, ratio test inconclusive, root test inconclusive"
– Agenog
Jan 23 '13 at 14:33
1) and 2) are obvious; 3) it's proved with some passages and with passage to the limit. But wolfram said: "sum does not converge, ratio test inconclusive, root test inconclusive"
– Agenog
Jan 23 '13 at 14:33
1
1
How can you prove (3) by passing to the limit, when you are trying to prove it for specific values, or do you just mean the function is strictly dereasing as a continuous function?
– Thomas Andrews
Jan 23 '13 at 14:36
How can you prove (3) by passing to the limit, when you are trying to prove it for specific values, or do you just mean the function is strictly dereasing as a continuous function?
– Thomas Andrews
Jan 23 '13 at 14:36
you can arrive to $log^2(n)> n*log^2(1+1/n) and with passage to limit the first part --> +inf, the second part of inequality -->0 so definitively I have proved the inequality
– Agenog
Jan 23 '13 at 14:58
you can arrive to $log^2(n)> n*log^2(1+1/n) and with passage to limit the first part --> +inf, the second part of inequality -->0 so definitively I have proved the inequality
– Agenog
Jan 23 '13 at 14:58
1
1
Again, you say "passage to the limit." Part of it is that you misformatted that comment, but I also just don't understand the thrust of your argument. What limit? If you send $ntoinfty$ how are you showing that $log(n)/sqrt n >log(n+1)/sqrtn+1$?
– Thomas Andrews
Jan 23 '13 at 15:18
Again, you say "passage to the limit." Part of it is that you misformatted that comment, but I also just don't understand the thrust of your argument. What limit? If you send $ntoinfty$ how are you showing that $log(n)/sqrt n >log(n+1)/sqrtn+1$?
– Thomas Andrews
Jan 23 '13 at 15:18
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
As you said 1. is obvious.
For 2., by De L'Hospital,
$$lim_nto +inftyfraclog nsqrt n=lim_nto +inftyfracfrac 1nfrac12sqrt n=lim_nto +inftyfrac2sqrt nn=0$$
For 3. you can either proceed with induction or show $f(x)=fraclog xsqrtx$ is stricly decreasing in $(N,+infty)$ (choose $N$ sufficiently large). Indeed,
$$f'(x)=fracfrac1 xsqrtx-fraclog x2sqrtxx$$
and
$$frac1 xsqrtx-fraclog x2sqrtx=frac2-log xsqrtx<0$$
for $x>e^2$
answered Jan 23 '13 at 14:36
Nameless
10k11753
Yes, then the series converges, right?
– Agenog
Jan 23 '13 at 14:50
@Giacomo By the alternating series (Leibniz) test, it does
– Nameless
Jan 23 '13 at 14:52
ok, thank you. But why wolfram said "does not converge" ?
– Agenog
Jan 23 '13 at 14:53
@Agenog Wolfram Alpha says it converges.
– user21467
Apr 5 '14 at 17:21
@Agenog I also read the answer but I didn't see the conclusion. It happens frequently in MSE. Anyway, the answer is pretty fine.
– Felix Marin
Sep 27 '16 at 21:17
add a comment |Â
up vote
3
down vote
$newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
Indeed, it can be explicitly evaluated such that a 'closed expression' exists.
Note that:
- beginalign
sum_n = 1^inftypars-1^n over n^s & =
pars2^s - 2zetaparss - 2^-sqquad
pars~zeta: Zeta Function~
endalign - Derive both members respect of $dss$:
beginalign
-sum_n = 1^inftypars-1^nlnparsn over n^s & =
-lnpars2zetaparss + lnpars2pars1 - 2^1 - szetaparss -
pars1 - 2^1 - szeta'parss
endalign - Take the limit $dss to 1/2$:
beginalign
color#f00sum_n = 1^inftypars-1^nlnparsn over rootn & =
root2lnpars2zetapars1 over 2 -
parsroot2 - 1zeta'pars1 over 2
\[5mm] & =
color#f00bracesroot2lnpars2 - 1 over 4parsroot2 - 1
bracksvphantomLarge A2gamma + pi +
2lnpars8pizetapars1 over 2
\[5mm] & approx 0.1933qquad
pars~gamma: Euler!-!Mascheroni Constant~
endalign
answered Sep 27 '16 at 21:41
Felix Marin
65.9k7105136
add a comment |Â
up vote
1
down vote
Yes, your solution is correct.
An alternative solution using the mean value theorem:
Note that if $a_n to_nto infty 0$ then $sum_n ge 0 (-1)^n a_n$ and $sum_n ge 0 (a_2n - a_2n-1)$ either both converge or both diverge. Set $a_n=frac ln nn^3/2$.
By the mean value theorem $a_2n - a_2n-1= frac2-log(t)t^3/2$ for some $t in ]2n-1;2n[$.
Now notice that because $fraclog qq^1/4 to 0$ there exists $q_0$ such that $(forall q>q_0),(ln q<q^1/4)$ .
Therefore $|frac2-log(t)t^3/2| le frac2+log(t)t^3/2 le frac2t^3/2 +frac1t^5/4 $.
Having this, it is simple to determine the convergence of the original problem.
answered Apr 5 '14 at 17:17
Marko Karbevski
1,035821
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
As you said 1. is obvious.
For 2., by De L'Hospital,
$$lim_nto +inftyfraclog nsqrt n=lim_nto +inftyfracfrac 1nfrac12sqrt n=lim_nto +inftyfrac2sqrt nn=0$$
For 3. you can either proceed with induction or show $f(x)=fraclog xsqrtx$ is stricly decreasing in $(N,+infty)$ (choose $N$ sufficiently large). Indeed,
$$f'(x)=fracfrac1 xsqrtx-fraclog x2sqrtxx$$
and
$$frac1 xsqrtx-fraclog x2sqrtx=frac2-log xsqrtx<0$$
for $x>e^2$
answered Jan 23 '13 at 14:36
Nameless
10k11753
Yes, then the series converges, right?
– Agenog
Jan 23 '13 at 14:50
@Giacomo By the alternating series (Leibniz) test, it does
– Nameless
Jan 23 '13 at 14:52
ok, thank you. But why wolfram said "does not converge" ?
– Agenog
Jan 23 '13 at 14:53
@Agenog Wolfram Alpha says it converges.
– user21467
Apr 5 '14 at 17:21
@Agenog I also read the answer but I didn't see the conclusion. It happens frequently in MSE. Anyway, the answer is pretty fine.
– Felix Marin
Sep 27 '16 at 21:17
add a comment |Â
up vote
5
down vote
accepted
As you said 1. is obvious.
For 2., by De L'Hospital,
$$lim_nto +inftyfraclog nsqrt n=lim_nto +inftyfracfrac 1nfrac12sqrt n=lim_nto +inftyfrac2sqrt nn=0$$
For 3. you can either proceed with induction or show $f(x)=fraclog xsqrtx$ is stricly decreasing in $(N,+infty)$ (choose $N$ sufficiently large). Indeed,
$$f'(x)=fracfrac1 xsqrtx-fraclog x2sqrtxx$$
and
$$frac1 xsqrtx-fraclog x2sqrtx=frac2-log xsqrtx<0$$
for $x>e^2$
answered Jan 23 '13 at 14:36
Nameless
10k11753
Yes, then the series converges, right?
– Agenog
Jan 23 '13 at 14:50
@Giacomo By the alternating series (Leibniz) test, it does
– Nameless
Jan 23 '13 at 14:52
ok, thank you. But why wolfram said "does not converge" ?
– Agenog
Jan 23 '13 at 14:53
@Agenog Wolfram Alpha says it converges.
– user21467
Apr 5 '14 at 17:21
@Agenog I also read the answer but I didn't see the conclusion. It happens frequently in MSE. Anyway, the answer is pretty fine.
– Felix Marin
Sep 27 '16 at 21:17
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
As you said 1. is obvious.
For 2., by De L'Hospital,
$$lim_nto +inftyfraclog nsqrt n=lim_nto +inftyfracfrac 1nfrac12sqrt n=lim_nto +inftyfrac2sqrt nn=0$$
For 3. you can either proceed with induction or show $f(x)=fraclog xsqrtx$ is stricly decreasing in $(N,+infty)$ (choose $N$ sufficiently large). Indeed,
$$f'(x)=fracfrac1 xsqrtx-fraclog x2sqrtxx$$
and
$$frac1 xsqrtx-fraclog x2sqrtx=frac2-log xsqrtx<0$$
for $x>e^2$
answered Jan 23 '13 at 14:36
Nameless
10k11753
As you said 1. is obvious.
For 2., by De L'Hospital,
$$lim_nto +inftyfraclog nsqrt n=lim_nto +inftyfracfrac 1nfrac12sqrt n=lim_nto +inftyfrac2sqrt nn=0$$
For 3. you can either proceed with induction or show $f(x)=fraclog xsqrtx$ is stricly decreasing in $(N,+infty)$ (choose $N$ sufficiently large). Indeed,
$$f'(x)=fracfrac1 xsqrtx-fraclog x2sqrtxx$$
and
$$frac1 xsqrtx-fraclog x2sqrtx=frac2-log xsqrtx<0$$
for $x>e^2$
answered Jan 23 '13 at 14:36
Nameless
10k11753
edited Jan 23 '13 at 15:02
answered Jan 23 '13 at 14:36
Nameless
10k11753
answered Jan 23 '13 at 14:36
Nameless
10k11753
answered Jan 23 '13 at 14:36
Nameless
10k11753
10k11753
Yes, then the series converges, right?
– Agenog
Jan 23 '13 at 14:50
@Giacomo By the alternating series (Leibniz) test, it does
– Nameless
Jan 23 '13 at 14:52
ok, thank you. But why wolfram said "does not converge" ?
– Agenog
Jan 23 '13 at 14:53
@Agenog Wolfram Alpha says it converges.
– user21467
Apr 5 '14 at 17:21
@Agenog I also read the answer but I didn't see the conclusion. It happens frequently in MSE. Anyway, the answer is pretty fine.
– Felix Marin
Sep 27 '16 at 21:17
add a comment |Â
Yes, then the series converges, right?
– Agenog
Jan 23 '13 at 14:50
@Giacomo By the alternating series (Leibniz) test, it does
– Nameless
Jan 23 '13 at 14:52
ok, thank you. But why wolfram said "does not converge" ?
– Agenog
Jan 23 '13 at 14:53
@Agenog Wolfram Alpha says it converges.
– user21467
Apr 5 '14 at 17:21
@Agenog I also read the answer but I didn't see the conclusion. It happens frequently in MSE. Anyway, the answer is pretty fine.
– Felix Marin
Sep 27 '16 at 21:17
Yes, then the series converges, right?
– Agenog
Jan 23 '13 at 14:50
Yes, then the series converges, right?
– Agenog
Jan 23 '13 at 14:50
@Giacomo By the alternating series (Leibniz) test, it does
– Nameless
Jan 23 '13 at 14:52
@Giacomo By the alternating series (Leibniz) test, it does
– Nameless
Jan 23 '13 at 14:52
ok, thank you. But why wolfram said "does not converge" ?
– Agenog
Jan 23 '13 at 14:53
ok, thank you. But why wolfram said "does not converge" ?
– Agenog
Jan 23 '13 at 14:53
@Agenog Wolfram Alpha says it converges.
– user21467
Apr 5 '14 at 17:21
@Agenog Wolfram Alpha says it converges.
– user21467
Apr 5 '14 at 17:21
@Agenog I also read the answer but I didn't see the conclusion. It happens frequently in MSE. Anyway, the answer is pretty fine.
– Felix Marin
Sep 27 '16 at 21:17
@Agenog I also read the answer but I didn't see the conclusion. It happens frequently in MSE. Anyway, the answer is pretty fine.
– Felix Marin
Sep 27 '16 at 21:17
add a comment |Â
up vote
3
down vote
$newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
Indeed, it can be explicitly evaluated such that a 'closed expression' exists.
Note that:
- beginalign
sum_n = 1^inftypars-1^n over n^s & =
pars2^s - 2zetaparss - 2^-sqquad
pars~zeta: Zeta Function~
endalign - Derive both members respect of $dss$:
beginalign
-sum_n = 1^inftypars-1^nlnparsn over n^s & =
-lnpars2zetaparss + lnpars2pars1 - 2^1 - szetaparss -
pars1 - 2^1 - szeta'parss
endalign - Take the limit $dss to 1/2$:
beginalign
color#f00sum_n = 1^inftypars-1^nlnparsn over rootn & =
root2lnpars2zetapars1 over 2 -
parsroot2 - 1zeta'pars1 over 2
\[5mm] & =
color#f00bracesroot2lnpars2 - 1 over 4parsroot2 - 1
bracksvphantomLarge A2gamma + pi +
2lnpars8pizetapars1 over 2
\[5mm] & approx 0.1933qquad
pars~gamma: Euler!-!Mascheroni Constant~
endalign
answered Sep 27 '16 at 21:41
Felix Marin
65.9k7105136
add a comment |Â
up vote
3
down vote
$newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
Indeed, it can be explicitly evaluated such that a 'closed expression' exists.
Note that:
- beginalign
sum_n = 1^inftypars-1^n over n^s & =
pars2^s - 2zetaparss - 2^-sqquad
pars~zeta: Zeta Function~
endalign - Derive both members respect of $dss$:
beginalign
-sum_n = 1^inftypars-1^nlnparsn over n^s & =
-lnpars2zetaparss + lnpars2pars1 - 2^1 - szetaparss -
pars1 - 2^1 - szeta'parss
endalign - Take the limit $dss to 1/2$:
beginalign
color#f00sum_n = 1^inftypars-1^nlnparsn over rootn & =
root2lnpars2zetapars1 over 2 -
parsroot2 - 1zeta'pars1 over 2
\[5mm] & =
color#f00bracesroot2lnpars2 - 1 over 4parsroot2 - 1
bracksvphantomLarge A2gamma + pi +
2lnpars8pizetapars1 over 2
\[5mm] & approx 0.1933qquad
pars~gamma: Euler!-!Mascheroni Constant~
endalign
answered Sep 27 '16 at 21:41
Felix Marin
65.9k7105136
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
Indeed, it can be explicitly evaluated such that a 'closed expression' exists.
Note that:
- beginalign
sum_n = 1^inftypars-1^n over n^s & =
pars2^s - 2zetaparss - 2^-sqquad
pars~zeta: Zeta Function~
endalign - Derive both members respect of $dss$:
beginalign
-sum_n = 1^inftypars-1^nlnparsn over n^s & =
-lnpars2zetaparss + lnpars2pars1 - 2^1 - szetaparss -
pars1 - 2^1 - szeta'parss
endalign - Take the limit $dss to 1/2$:
beginalign
color#f00sum_n = 1^inftypars-1^nlnparsn over rootn & =
root2lnpars2zetapars1 over 2 -
parsroot2 - 1zeta'pars1 over 2
\[5mm] & =
color#f00bracesroot2lnpars2 - 1 over 4parsroot2 - 1
bracksvphantomLarge A2gamma + pi +
2lnpars8pizetapars1 over 2
\[5mm] & approx 0.1933qquad
pars~gamma: Euler!-!Mascheroni Constant~
endalign
answered Sep 27 '16 at 21:41
Felix Marin
65.9k7105136
$newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
Indeed, it can be explicitly evaluated such that a 'closed expression' exists.
Note that:
- beginalign
sum_n = 1^inftypars-1^n over n^s & =
pars2^s - 2zetaparss - 2^-sqquad
pars~zeta: Zeta Function~
endalign - Derive both members respect of $dss$:
beginalign
-sum_n = 1^inftypars-1^nlnparsn over n^s & =
-lnpars2zetaparss + lnpars2pars1 - 2^1 - szetaparss -
pars1 - 2^1 - szeta'parss
endalign - Take the limit $dss to 1/2$:
beginalign
color#f00sum_n = 1^inftypars-1^nlnparsn over rootn & =
root2lnpars2zetapars1 over 2 -
parsroot2 - 1zeta'pars1 over 2
\[5mm] & =
color#f00bracesroot2lnpars2 - 1 over 4parsroot2 - 1
bracksvphantomLarge A2gamma + pi +
2lnpars8pizetapars1 over 2
\[5mm] & approx 0.1933qquad
pars~gamma: Euler!-!Mascheroni Constant~
endalign
answered Sep 27 '16 at 21:41
Felix Marin
65.9k7105136
edited Sep 27 '16 at 23:13
answered Sep 27 '16 at 21:41
Felix Marin
65.9k7105136
answered Sep 27 '16 at 21:41
Felix Marin
65.9k7105136
answered Sep 27 '16 at 21:41
Felix Marin
65.9k7105136
65.9k7105136
add a comment |Â
add a comment |Â
up vote
1
down vote
Yes, your solution is correct.
An alternative solution using the mean value theorem:
Note that if $a_n to_nto infty 0$ then $sum_n ge 0 (-1)^n a_n$ and $sum_n ge 0 (a_2n - a_2n-1)$ either both converge or both diverge. Set $a_n=frac ln nn^3/2$.
By the mean value theorem $a_2n - a_2n-1= frac2-log(t)t^3/2$ for some $t in ]2n-1;2n[$.
Now notice that because $fraclog qq^1/4 to 0$ there exists $q_0$ such that $(forall q>q_0),(ln q<q^1/4)$ .
Therefore $|frac2-log(t)t^3/2| le frac2+log(t)t^3/2 le frac2t^3/2 +frac1t^5/4 $.
Having this, it is simple to determine the convergence of the original problem.
answered Apr 5 '14 at 17:17
Marko Karbevski
1,035821
add a comment |Â
up vote
1
down vote
Yes, your solution is correct.
An alternative solution using the mean value theorem:
Note that if $a_n to_nto infty 0$ then $sum_n ge 0 (-1)^n a_n$ and $sum_n ge 0 (a_2n - a_2n-1)$ either both converge or both diverge. Set $a_n=frac ln nn^3/2$.
By the mean value theorem $a_2n - a_2n-1= frac2-log(t)t^3/2$ for some $t in ]2n-1;2n[$.
Now notice that because $fraclog qq^1/4 to 0$ there exists $q_0$ such that $(forall q>q_0),(ln q<q^1/4)$ .
Therefore $|frac2-log(t)t^3/2| le frac2+log(t)t^3/2 le frac2t^3/2 +frac1t^5/4 $.
Having this, it is simple to determine the convergence of the original problem.
answered Apr 5 '14 at 17:17
Marko Karbevski
1,035821
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, your solution is correct.
An alternative solution using the mean value theorem:
Note that if $a_n to_nto infty 0$ then $sum_n ge 0 (-1)^n a_n$ and $sum_n ge 0 (a_2n - a_2n-1)$ either both converge or both diverge. Set $a_n=frac ln nn^3/2$.
By the mean value theorem $a_2n - a_2n-1= frac2-log(t)t^3/2$ for some $t in ]2n-1;2n[$.
Now notice that because $fraclog qq^1/4 to 0$ there exists $q_0$ such that $(forall q>q_0),(ln q<q^1/4)$ .
Therefore $|frac2-log(t)t^3/2| le frac2+log(t)t^3/2 le frac2t^3/2 +frac1t^5/4 $.
Having this, it is simple to determine the convergence of the original problem.
answered Apr 5 '14 at 17:17
Marko Karbevski
1,035821
Yes, your solution is correct.
An alternative solution using the mean value theorem:
Note that if $a_n to_nto infty 0$ then $sum_n ge 0 (-1)^n a_n$ and $sum_n ge 0 (a_2n - a_2n-1)$ either both converge or both diverge. Set $a_n=frac ln nn^3/2$.
By the mean value theorem $a_2n - a_2n-1= frac2-log(t)t^3/2$ for some $t in ]2n-1;2n[$.
Now notice that because $fraclog qq^1/4 to 0$ there exists $q_0$ such that $(forall q>q_0),(ln q<q^1/4)$ .
Therefore $|frac2-log(t)t^3/2| le frac2+log(t)t^3/2 le frac2t^3/2 +frac1t^5/4 $.
Having this, it is simple to determine the convergence of the original problem.
answered Apr 5 '14 at 17:17
Marko Karbevski
1,035821
edited Sep 27 '16 at 20:51
answered Apr 5 '14 at 17:17
Marko Karbevski
1,035821
answered Apr 5 '14 at 17:17
Marko Karbevski
1,035821
answered Apr 5 '14 at 17:17
Marko Karbevski
1,035821
1,035821
add a comment |Â
add a comment |Â
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That is what you need to show, yes. Best to be explicit about what $a_n$ is from the start. You obviously can't just assert (2) and (3), but have to prove them.
– Thomas Andrews
Jan 23 '13 at 14:28
1) and 2) are obvious; 3) it's proved with some passages and with passage to the limit. But wolfram said: "sum does not converge, ratio test inconclusive, root test inconclusive"
– Agenog
Jan 23 '13 at 14:33
1
How can you prove (3) by passing to the limit, when you are trying to prove it for specific values, or do you just mean the function is strictly dereasing as a continuous function?
– Thomas Andrews
Jan 23 '13 at 14:36
you can arrive to $log^2(n)> n*log^2(1+1/n) and with passage to limit the first part --> +inf, the second part of inequality -->0 so definitively I have proved the inequality
– Agenog
Jan 23 '13 at 14:58
1
Again, you say "passage to the limit." Part of it is that you misformatted that comment, but I also just don't understand the thrust of your argument. What limit? If you send $ntoinfty$ how are you showing that $log(n)/sqrt n >log(n+1)/sqrtn+1$?
– Thomas Andrews
Jan 23 '13 at 15:18