Convergency of $sum_n=1^infty fracfrac12+(-1)^nn$
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0
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Does the following series converge?: $$sum_n=1^infty fracfrac12+(-1)^nn$$
Liebnitz test shouldn't work here...Also I try to find out the partial sum sequence but it was not so fruitful ....Please help.
real-analysis sequences-and-series
asked Aug 23 at 2:39
Indrajit Ghosh
763515
add a comment |Â
up vote
0
down vote
favorite
Does the following series converge?: $$sum_n=1^infty fracfrac12+(-1)^nn$$
Liebnitz test shouldn't work here...Also I try to find out the partial sum sequence but it was not so fruitful ....Please help.
real-analysis sequences-and-series
asked Aug 23 at 2:39
Indrajit Ghosh
763515
$frac12+(-1)^n$ has a positive mean value hence your series is divergent by Kronecker's lemma.
– Jack D'Aurizio♦
Aug 23 at 14:51
What is Kronecker Lemma..?
– Indrajit Ghosh
Aug 23 at 14:52
en.wikipedia.org/wiki/Kronecker%27s_lemma
– Jack D'Aurizio♦
Aug 23 at 15:01
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does the following series converge?: $$sum_n=1^infty fracfrac12+(-1)^nn$$
Liebnitz test shouldn't work here...Also I try to find out the partial sum sequence but it was not so fruitful ....Please help.
real-analysis sequences-and-series
asked Aug 23 at 2:39
Indrajit Ghosh
763515
Does the following series converge?: $$sum_n=1^infty fracfrac12+(-1)^nn$$
Liebnitz test shouldn't work here...Also I try to find out the partial sum sequence but it was not so fruitful ....Please help.
real-analysis sequences-and-series
asked Aug 23 at 2:39
Indrajit Ghosh
763515
asked Aug 23 at 2:39
Indrajit Ghosh
763515
asked Aug 23 at 2:39
Indrajit Ghosh
763515
asked Aug 23 at 2:39
Indrajit Ghosh
763515
763515
$frac12+(-1)^n$ has a positive mean value hence your series is divergent by Kronecker's lemma.
– Jack D'Aurizio♦
Aug 23 at 14:51
What is Kronecker Lemma..?
– Indrajit Ghosh
Aug 23 at 14:52
en.wikipedia.org/wiki/Kronecker%27s_lemma
– Jack D'Aurizio♦
Aug 23 at 15:01
add a comment |Â
$frac12+(-1)^n$ has a positive mean value hence your series is divergent by Kronecker's lemma.
– Jack D'Aurizio♦
Aug 23 at 14:51
What is Kronecker Lemma..?
– Indrajit Ghosh
Aug 23 at 14:52
en.wikipedia.org/wiki/Kronecker%27s_lemma
– Jack D'Aurizio♦
Aug 23 at 15:01
$frac12+(-1)^n$ has a positive mean value hence your series is divergent by Kronecker's lemma.
– Jack D'Aurizio♦
Aug 23 at 14:51
$frac12+(-1)^n$ has a positive mean value hence your series is divergent by Kronecker's lemma.
– Jack D'Aurizio♦
Aug 23 at 14:51
What is Kronecker Lemma..?
– Indrajit Ghosh
Aug 23 at 14:52
What is Kronecker Lemma..?
– Indrajit Ghosh
Aug 23 at 14:52
en.wikipedia.org/wiki/Kronecker%27s_lemma
– Jack D'Aurizio♦
Aug 23 at 15:01
en.wikipedia.org/wiki/Kronecker%27s_lemma
– Jack D'Aurizio♦
Aug 23 at 15:01
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Hint
$$S_n= left(sum_k=1^n frac12k right)+left(sum_k=1^n frac(-1)^kk right)$$
The first series diverges and the second converges by Leibnitz test...
answered Aug 23 at 2:43
N. S.
98.5k5106197
But unless we prove the absolute convergency we cannot rearrange its terms..
– Indrajit Ghosh
Aug 23 at 2:49
1
@IndrajitGhosh There is no rearrangement here.
– Mark Viola
Aug 23 at 3:00
Yes. Divergence means not convergent. Then, since we have a more explicit method of proving convergence, we should assume convergence and deduce contradiction. Assume a convergent sequence $A_n$ and divergent one $B_n$, with sums $A$ and $B$, summed together is convergent.(Let the summation of A_n+B_n be C). Now, -A_n is convergent for $A_n$ is(note $lim_n to infty -A_n=-lim_n to infty A_n$, or $-A$. Then, $-A+A+B$ would be the sum of a convergent and convergent sequence. This would mean, though, $B$ is convergent, a contradiction.
– VgAcid
Aug 23 at 3:23
@IndrajitGhosh Yes. And since we are talking about the partial sums, they are finite and you can rearrange.... The issues with rearranging series is that by rearranging, you change the partial sums.
– N. S.
Aug 23 at 3:24
1
@IndrajitGhosh The theorem I reffer to states: If $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ are convergent, then so is $sum_n=1^infty (a_n-b_n)$ and $$sum_n=1^infty ( a_n- b_n)= (sum_n=1^infty a_n)-(sum_n=1^infty b_n)$$
– N. S.
Aug 23 at 3:33
 |Â
show 2 more comments
up vote
1
down vote
Since
$$|frac12+(-1)^n|geq|(-1)^n|-frac12=frac12$$
then
$$sum_n=1^infty |fracfrac12+(-1)^nn|>dfrac12sum_n=1dfrac1n$$
answered Aug 23 at 2:53
Nosrati
21.1k41645
Uhm... But some terms are still negative, so what do you make out of that inequality?
– Saucy O'Path
Aug 23 at 2:56
1
But it doesn't mean the series is divergent....it may be conditionally convergent.
– Indrajit Ghosh
Aug 23 at 2:56
this shows the series isn't absolutely converge!
– Nosrati
Aug 23 at 2:57
@Nosrati The other answer does show that the series diverges.
– Mark Viola
Aug 23 at 3:01
@Nosrati The first sequence in $S_n$ is divergent and the 2nd one converges to $log 2$..does it mean $S_n$ is also divergent?
– Indrajit Ghosh
Aug 23 at 3:12
 |Â
show 2 more comments
up vote
1
down vote
Let $A(m)=frac frac 12+(-1)^2m2m+$ $frac frac 12+(-1)^2m+12m+1.$
We have $A(m)=frac 3/22m-frac 1/22m+1>$ $frac 3/22m-$ $frac 1/22m=$ $frac 12m.$
Therefore $sum_n=1^2M+1left(frac frac 12+(-1)^nnright)=$ $-1/2+sum_m=1^M A(m)>$ $-1/2+sum_m=1^Mfrac 12m$
which $to infty$ as $Mto infty.$
answered Aug 23 at 5:28
DanielWainfleet
32k31644
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint
$$S_n= left(sum_k=1^n frac12k right)+left(sum_k=1^n frac(-1)^kk right)$$
The first series diverges and the second converges by Leibnitz test...
answered Aug 23 at 2:43
N. S.
98.5k5106197
But unless we prove the absolute convergency we cannot rearrange its terms..
– Indrajit Ghosh
Aug 23 at 2:49
1
@IndrajitGhosh There is no rearrangement here.
– Mark Viola
Aug 23 at 3:00
Yes. Divergence means not convergent. Then, since we have a more explicit method of proving convergence, we should assume convergence and deduce contradiction. Assume a convergent sequence $A_n$ and divergent one $B_n$, with sums $A$ and $B$, summed together is convergent.(Let the summation of A_n+B_n be C). Now, -A_n is convergent for $A_n$ is(note $lim_n to infty -A_n=-lim_n to infty A_n$, or $-A$. Then, $-A+A+B$ would be the sum of a convergent and convergent sequence. This would mean, though, $B$ is convergent, a contradiction.
– VgAcid
Aug 23 at 3:23
@IndrajitGhosh Yes. And since we are talking about the partial sums, they are finite and you can rearrange.... The issues with rearranging series is that by rearranging, you change the partial sums.
– N. S.
Aug 23 at 3:24
1
@IndrajitGhosh The theorem I reffer to states: If $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ are convergent, then so is $sum_n=1^infty (a_n-b_n)$ and $$sum_n=1^infty ( a_n- b_n)= (sum_n=1^infty a_n)-(sum_n=1^infty b_n)$$
– N. S.
Aug 23 at 3:33
 |Â
show 2 more comments
up vote
2
down vote
accepted
Hint
$$S_n= left(sum_k=1^n frac12k right)+left(sum_k=1^n frac(-1)^kk right)$$
The first series diverges and the second converges by Leibnitz test...
answered Aug 23 at 2:43
N. S.
98.5k5106197
But unless we prove the absolute convergency we cannot rearrange its terms..
– Indrajit Ghosh
Aug 23 at 2:49
1
@IndrajitGhosh There is no rearrangement here.
– Mark Viola
Aug 23 at 3:00
Yes. Divergence means not convergent. Then, since we have a more explicit method of proving convergence, we should assume convergence and deduce contradiction. Assume a convergent sequence $A_n$ and divergent one $B_n$, with sums $A$ and $B$, summed together is convergent.(Let the summation of A_n+B_n be C). Now, -A_n is convergent for $A_n$ is(note $lim_n to infty -A_n=-lim_n to infty A_n$, or $-A$. Then, $-A+A+B$ would be the sum of a convergent and convergent sequence. This would mean, though, $B$ is convergent, a contradiction.
– VgAcid
Aug 23 at 3:23
@IndrajitGhosh Yes. And since we are talking about the partial sums, they are finite and you can rearrange.... The issues with rearranging series is that by rearranging, you change the partial sums.
– N. S.
Aug 23 at 3:24
1
@IndrajitGhosh The theorem I reffer to states: If $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ are convergent, then so is $sum_n=1^infty (a_n-b_n)$ and $$sum_n=1^infty ( a_n- b_n)= (sum_n=1^infty a_n)-(sum_n=1^infty b_n)$$
– N. S.
Aug 23 at 3:33
 |Â
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint
$$S_n= left(sum_k=1^n frac12k right)+left(sum_k=1^n frac(-1)^kk right)$$
The first series diverges and the second converges by Leibnitz test...
answered Aug 23 at 2:43
N. S.
98.5k5106197
Hint
$$S_n= left(sum_k=1^n frac12k right)+left(sum_k=1^n frac(-1)^kk right)$$
The first series diverges and the second converges by Leibnitz test...
answered Aug 23 at 2:43
N. S.
98.5k5106197
answered Aug 23 at 2:43
N. S.
98.5k5106197
answered Aug 23 at 2:43
N. S.
98.5k5106197
answered Aug 23 at 2:43
N. S.
98.5k5106197
98.5k5106197
But unless we prove the absolute convergency we cannot rearrange its terms..
– Indrajit Ghosh
Aug 23 at 2:49
1
@IndrajitGhosh There is no rearrangement here.
– Mark Viola
Aug 23 at 3:00
Yes. Divergence means not convergent. Then, since we have a more explicit method of proving convergence, we should assume convergence and deduce contradiction. Assume a convergent sequence $A_n$ and divergent one $B_n$, with sums $A$ and $B$, summed together is convergent.(Let the summation of A_n+B_n be C). Now, -A_n is convergent for $A_n$ is(note $lim_n to infty -A_n=-lim_n to infty A_n$, or $-A$. Then, $-A+A+B$ would be the sum of a convergent and convergent sequence. This would mean, though, $B$ is convergent, a contradiction.
– VgAcid
Aug 23 at 3:23
@IndrajitGhosh Yes. And since we are talking about the partial sums, they are finite and you can rearrange.... The issues with rearranging series is that by rearranging, you change the partial sums.
– N. S.
Aug 23 at 3:24
1
@IndrajitGhosh The theorem I reffer to states: If $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ are convergent, then so is $sum_n=1^infty (a_n-b_n)$ and $$sum_n=1^infty ( a_n- b_n)= (sum_n=1^infty a_n)-(sum_n=1^infty b_n)$$
– N. S.
Aug 23 at 3:33
 |Â
show 2 more comments
But unless we prove the absolute convergency we cannot rearrange its terms..
– Indrajit Ghosh
Aug 23 at 2:49
1
@IndrajitGhosh There is no rearrangement here.
– Mark Viola
Aug 23 at 3:00
Yes. Divergence means not convergent. Then, since we have a more explicit method of proving convergence, we should assume convergence and deduce contradiction. Assume a convergent sequence $A_n$ and divergent one $B_n$, with sums $A$ and $B$, summed together is convergent.(Let the summation of A_n+B_n be C). Now, -A_n is convergent for $A_n$ is(note $lim_n to infty -A_n=-lim_n to infty A_n$, or $-A$. Then, $-A+A+B$ would be the sum of a convergent and convergent sequence. This would mean, though, $B$ is convergent, a contradiction.
– VgAcid
Aug 23 at 3:23
@IndrajitGhosh Yes. And since we are talking about the partial sums, they are finite and you can rearrange.... The issues with rearranging series is that by rearranging, you change the partial sums.
– N. S.
Aug 23 at 3:24
1
@IndrajitGhosh The theorem I reffer to states: If $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ are convergent, then so is $sum_n=1^infty (a_n-b_n)$ and $$sum_n=1^infty ( a_n- b_n)= (sum_n=1^infty a_n)-(sum_n=1^infty b_n)$$
– N. S.
Aug 23 at 3:33
But unless we prove the absolute convergency we cannot rearrange its terms..
– Indrajit Ghosh
Aug 23 at 2:49
But unless we prove the absolute convergency we cannot rearrange its terms..
– Indrajit Ghosh
Aug 23 at 2:49
1
1
@IndrajitGhosh There is no rearrangement here.
– Mark Viola
Aug 23 at 3:00
@IndrajitGhosh There is no rearrangement here.
– Mark Viola
Aug 23 at 3:00
Yes. Divergence means not convergent. Then, since we have a more explicit method of proving convergence, we should assume convergence and deduce contradiction. Assume a convergent sequence $A_n$ and divergent one $B_n$, with sums $A$ and $B$, summed together is convergent.(Let the summation of A_n+B_n be C). Now, -A_n is convergent for $A_n$ is(note $lim_n to infty -A_n=-lim_n to infty A_n$, or $-A$. Then, $-A+A+B$ would be the sum of a convergent and convergent sequence. This would mean, though, $B$ is convergent, a contradiction.
– VgAcid
Aug 23 at 3:23
Yes. Divergence means not convergent. Then, since we have a more explicit method of proving convergence, we should assume convergence and deduce contradiction. Assume a convergent sequence $A_n$ and divergent one $B_n$, with sums $A$ and $B$, summed together is convergent.(Let the summation of A_n+B_n be C). Now, -A_n is convergent for $A_n$ is(note $lim_n to infty -A_n=-lim_n to infty A_n$, or $-A$. Then, $-A+A+B$ would be the sum of a convergent and convergent sequence. This would mean, though, $B$ is convergent, a contradiction.
– VgAcid
Aug 23 at 3:23
@IndrajitGhosh Yes. And since we are talking about the partial sums, they are finite and you can rearrange.... The issues with rearranging series is that by rearranging, you change the partial sums.
– N. S.
Aug 23 at 3:24
@IndrajitGhosh Yes. And since we are talking about the partial sums, they are finite and you can rearrange.... The issues with rearranging series is that by rearranging, you change the partial sums.
– N. S.
Aug 23 at 3:24
1
1
@IndrajitGhosh The theorem I reffer to states: If $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ are convergent, then so is $sum_n=1^infty (a_n-b_n)$ and $$sum_n=1^infty ( a_n- b_n)= (sum_n=1^infty a_n)-(sum_n=1^infty b_n)$$
– N. S.
Aug 23 at 3:33
@IndrajitGhosh The theorem I reffer to states: If $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ are convergent, then so is $sum_n=1^infty (a_n-b_n)$ and $$sum_n=1^infty ( a_n- b_n)= (sum_n=1^infty a_n)-(sum_n=1^infty b_n)$$
– N. S.
Aug 23 at 3:33
 |Â
show 2 more comments
up vote
1
down vote
Since
$$|frac12+(-1)^n|geq|(-1)^n|-frac12=frac12$$
then
$$sum_n=1^infty |fracfrac12+(-1)^nn|>dfrac12sum_n=1dfrac1n$$
answered Aug 23 at 2:53
Nosrati
21.1k41645
Uhm... But some terms are still negative, so what do you make out of that inequality?
– Saucy O'Path
Aug 23 at 2:56
1
But it doesn't mean the series is divergent....it may be conditionally convergent.
– Indrajit Ghosh
Aug 23 at 2:56
this shows the series isn't absolutely converge!
– Nosrati
Aug 23 at 2:57
@Nosrati The other answer does show that the series diverges.
– Mark Viola
Aug 23 at 3:01
@Nosrati The first sequence in $S_n$ is divergent and the 2nd one converges to $log 2$..does it mean $S_n$ is also divergent?
– Indrajit Ghosh
Aug 23 at 3:12
 |Â
show 2 more comments
up vote
1
down vote
Since
$$|frac12+(-1)^n|geq|(-1)^n|-frac12=frac12$$
then
$$sum_n=1^infty |fracfrac12+(-1)^nn|>dfrac12sum_n=1dfrac1n$$
answered Aug 23 at 2:53
Nosrati
21.1k41645
Uhm... But some terms are still negative, so what do you make out of that inequality?
– Saucy O'Path
Aug 23 at 2:56
1
But it doesn't mean the series is divergent....it may be conditionally convergent.
– Indrajit Ghosh
Aug 23 at 2:56
this shows the series isn't absolutely converge!
– Nosrati
Aug 23 at 2:57
@Nosrati The other answer does show that the series diverges.
– Mark Viola
Aug 23 at 3:01
@Nosrati The first sequence in $S_n$ is divergent and the 2nd one converges to $log 2$..does it mean $S_n$ is also divergent?
– Indrajit Ghosh
Aug 23 at 3:12
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
Since
$$|frac12+(-1)^n|geq|(-1)^n|-frac12=frac12$$
then
$$sum_n=1^infty |fracfrac12+(-1)^nn|>dfrac12sum_n=1dfrac1n$$
answered Aug 23 at 2:53
Nosrati
21.1k41645
Since
$$|frac12+(-1)^n|geq|(-1)^n|-frac12=frac12$$
then
$$sum_n=1^infty |fracfrac12+(-1)^nn|>dfrac12sum_n=1dfrac1n$$
answered Aug 23 at 2:53
Nosrati
21.1k41645
answered Aug 23 at 2:53
Nosrati
21.1k41645
answered Aug 23 at 2:53
Nosrati
21.1k41645
answered Aug 23 at 2:53
Nosrati
21.1k41645
21.1k41645
Uhm... But some terms are still negative, so what do you make out of that inequality?
– Saucy O'Path
Aug 23 at 2:56
1
But it doesn't mean the series is divergent....it may be conditionally convergent.
– Indrajit Ghosh
Aug 23 at 2:56
this shows the series isn't absolutely converge!
– Nosrati
Aug 23 at 2:57
@Nosrati The other answer does show that the series diverges.
– Mark Viola
Aug 23 at 3:01
@Nosrati The first sequence in $S_n$ is divergent and the 2nd one converges to $log 2$..does it mean $S_n$ is also divergent?
– Indrajit Ghosh
Aug 23 at 3:12
 |Â
show 2 more comments
Uhm... But some terms are still negative, so what do you make out of that inequality?
– Saucy O'Path
Aug 23 at 2:56
1
But it doesn't mean the series is divergent....it may be conditionally convergent.
– Indrajit Ghosh
Aug 23 at 2:56
this shows the series isn't absolutely converge!
– Nosrati
Aug 23 at 2:57
@Nosrati The other answer does show that the series diverges.
– Mark Viola
Aug 23 at 3:01
@Nosrati The first sequence in $S_n$ is divergent and the 2nd one converges to $log 2$..does it mean $S_n$ is also divergent?
– Indrajit Ghosh
Aug 23 at 3:12
Uhm... But some terms are still negative, so what do you make out of that inequality?
– Saucy O'Path
Aug 23 at 2:56
Uhm... But some terms are still negative, so what do you make out of that inequality?
– Saucy O'Path
Aug 23 at 2:56
1
1
But it doesn't mean the series is divergent....it may be conditionally convergent.
– Indrajit Ghosh
Aug 23 at 2:56
But it doesn't mean the series is divergent....it may be conditionally convergent.
– Indrajit Ghosh
Aug 23 at 2:56
this shows the series isn't absolutely converge!
– Nosrati
Aug 23 at 2:57
this shows the series isn't absolutely converge!
– Nosrati
Aug 23 at 2:57
@Nosrati The other answer does show that the series diverges.
– Mark Viola
Aug 23 at 3:01
@Nosrati The other answer does show that the series diverges.
– Mark Viola
Aug 23 at 3:01
@Nosrati The first sequence in $S_n$ is divergent and the 2nd one converges to $log 2$..does it mean $S_n$ is also divergent?
– Indrajit Ghosh
Aug 23 at 3:12
@Nosrati The first sequence in $S_n$ is divergent and the 2nd one converges to $log 2$..does it mean $S_n$ is also divergent?
– Indrajit Ghosh
Aug 23 at 3:12
 |Â
show 2 more comments
up vote
1
down vote
Let $A(m)=frac frac 12+(-1)^2m2m+$ $frac frac 12+(-1)^2m+12m+1.$
We have $A(m)=frac 3/22m-frac 1/22m+1>$ $frac 3/22m-$ $frac 1/22m=$ $frac 12m.$
Therefore $sum_n=1^2M+1left(frac frac 12+(-1)^nnright)=$ $-1/2+sum_m=1^M A(m)>$ $-1/2+sum_m=1^Mfrac 12m$
which $to infty$ as $Mto infty.$
answered Aug 23 at 5:28
DanielWainfleet
32k31644
add a comment |Â
up vote
1
down vote
Let $A(m)=frac frac 12+(-1)^2m2m+$ $frac frac 12+(-1)^2m+12m+1.$
We have $A(m)=frac 3/22m-frac 1/22m+1>$ $frac 3/22m-$ $frac 1/22m=$ $frac 12m.$
Therefore $sum_n=1^2M+1left(frac frac 12+(-1)^nnright)=$ $-1/2+sum_m=1^M A(m)>$ $-1/2+sum_m=1^Mfrac 12m$
which $to infty$ as $Mto infty.$
answered Aug 23 at 5:28
DanielWainfleet
32k31644
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $A(m)=frac frac 12+(-1)^2m2m+$ $frac frac 12+(-1)^2m+12m+1.$
We have $A(m)=frac 3/22m-frac 1/22m+1>$ $frac 3/22m-$ $frac 1/22m=$ $frac 12m.$
Therefore $sum_n=1^2M+1left(frac frac 12+(-1)^nnright)=$ $-1/2+sum_m=1^M A(m)>$ $-1/2+sum_m=1^Mfrac 12m$
which $to infty$ as $Mto infty.$
answered Aug 23 at 5:28
DanielWainfleet
32k31644
Let $A(m)=frac frac 12+(-1)^2m2m+$ $frac frac 12+(-1)^2m+12m+1.$
We have $A(m)=frac 3/22m-frac 1/22m+1>$ $frac 3/22m-$ $frac 1/22m=$ $frac 12m.$
Therefore $sum_n=1^2M+1left(frac frac 12+(-1)^nnright)=$ $-1/2+sum_m=1^M A(m)>$ $-1/2+sum_m=1^Mfrac 12m$
which $to infty$ as $Mto infty.$
answered Aug 23 at 5:28
DanielWainfleet
32k31644
answered Aug 23 at 5:28
DanielWainfleet
32k31644
answered Aug 23 at 5:28
DanielWainfleet
32k31644
answered Aug 23 at 5:28
DanielWainfleet
32k31644
32k31644
add a comment |Â
add a comment |Â
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$frac12+(-1)^n$ has a positive mean value hence your series is divergent by Kronecker's lemma.
– Jack D'Aurizio♦
Aug 23 at 14:51
What is Kronecker Lemma..?
– Indrajit Ghosh
Aug 23 at 14:52
en.wikipedia.org/wiki/Kronecker%27s_lemma
– Jack D'Aurizio♦
Aug 23 at 15:01