Absolute convergence of series $sumlimits_n=2^infty(-1)^n-1over (n+(-1)^nsqrtn)^2over 3$
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Exam absolute convergence of series $$sumlimits_n=2^infty(-1)^n-1over (n+(-1)^nsqrtn)^2over 3$$
WolframAlpha says it diverges, but I don't know how to show it. As it's not monotonic $Rightarrow$ I can't estimate it from below. I tried to do áauchy criteria, but I got mess with radicals and got nothing
sequences-and-series convergence absolute-convergence
asked Dec 20 '17 at 19:32
ioleg19029700
444110
add a comment |Â
up vote
1
down vote
favorite
Exam absolute convergence of series $$sumlimits_n=2^infty(-1)^n-1over (n+(-1)^nsqrtn)^2over 3$$
WolframAlpha says it diverges, but I don't know how to show it. As it's not monotonic $Rightarrow$ I can't estimate it from below. I tried to do áauchy criteria, but I got mess with radicals and got nothing
sequences-and-series convergence absolute-convergence
asked Dec 20 '17 at 19:32
ioleg19029700
444110
Why not just do a limit comparison with $sum frac1{n^frac23 $?
– Rene Schipperus
Dec 20 '17 at 19:35
Absolute divergence is direct. Are you not asked for (non absolute) convergence?
– Did
Dec 20 '17 at 19:35
@ReneSchipperus, Limit Comparison requires nonnegativity of both the series. The OP's series is not nonnegative always.
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:38
@MrigankaBasuRoyChowdhury I thought he asked about absolute convergence ?
– Rene Schipperus
Dec 20 '17 at 19:40
@ReneSchipperus: Ohh, then its okay, my bad :(. But seeing how trivial it is, maybe he actually did not. :)
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:41
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Exam absolute convergence of series $$sumlimits_n=2^infty(-1)^n-1over (n+(-1)^nsqrtn)^2over 3$$
WolframAlpha says it diverges, but I don't know how to show it. As it's not monotonic $Rightarrow$ I can't estimate it from below. I tried to do áauchy criteria, but I got mess with radicals and got nothing
sequences-and-series convergence absolute-convergence
asked Dec 20 '17 at 19:32
ioleg19029700
444110
Exam absolute convergence of series $$sumlimits_n=2^infty(-1)^n-1over (n+(-1)^nsqrtn)^2over 3$$
WolframAlpha says it diverges, but I don't know how to show it. As it's not monotonic $Rightarrow$ I can't estimate it from below. I tried to do áauchy criteria, but I got mess with radicals and got nothing
sequences-and-series convergence absolute-convergence
asked Dec 20 '17 at 19:32
ioleg19029700
444110
asked Dec 20 '17 at 19:32
ioleg19029700
444110
asked Dec 20 '17 at 19:32
ioleg19029700
444110
asked Dec 20 '17 at 19:32
ioleg19029700
444110
444110
Why not just do a limit comparison with $sum frac1{n^frac23 $?
– Rene Schipperus
Dec 20 '17 at 19:35
Absolute divergence is direct. Are you not asked for (non absolute) convergence?
– Did
Dec 20 '17 at 19:35
@ReneSchipperus, Limit Comparison requires nonnegativity of both the series. The OP's series is not nonnegative always.
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:38
@MrigankaBasuRoyChowdhury I thought he asked about absolute convergence ?
– Rene Schipperus
Dec 20 '17 at 19:40
@ReneSchipperus: Ohh, then its okay, my bad :(. But seeing how trivial it is, maybe he actually did not. :)
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:41
add a comment |Â
Why not just do a limit comparison with $sum frac1{n^frac23 $?
– Rene Schipperus
Dec 20 '17 at 19:35
Absolute divergence is direct. Are you not asked for (non absolute) convergence?
– Did
Dec 20 '17 at 19:35
@ReneSchipperus, Limit Comparison requires nonnegativity of both the series. The OP's series is not nonnegative always.
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:38
@MrigankaBasuRoyChowdhury I thought he asked about absolute convergence ?
– Rene Schipperus
Dec 20 '17 at 19:40
@ReneSchipperus: Ohh, then its okay, my bad :(. But seeing how trivial it is, maybe he actually did not. :)
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:41
Why not just do a limit comparison with $sum frac1{n^frac23 $?
– Rene Schipperus
Dec 20 '17 at 19:35
Why not just do a limit comparison with $sum frac1{n^frac23 $?
– Rene Schipperus
Dec 20 '17 at 19:35
Absolute divergence is direct. Are you not asked for (non absolute) convergence?
– Did
Dec 20 '17 at 19:35
Absolute divergence is direct. Are you not asked for (non absolute) convergence?
– Did
Dec 20 '17 at 19:35
@ReneSchipperus, Limit Comparison requires nonnegativity of both the series. The OP's series is not nonnegative always.
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:38
@ReneSchipperus, Limit Comparison requires nonnegativity of both the series. The OP's series is not nonnegative always.
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:38
@MrigankaBasuRoyChowdhury I thought he asked about absolute convergence ?
– Rene Schipperus
Dec 20 '17 at 19:40
@MrigankaBasuRoyChowdhury I thought he asked about absolute convergence ?
– Rene Schipperus
Dec 20 '17 at 19:40
@ReneSchipperus: Ohh, then its okay, my bad :(. But seeing how trivial it is, maybe he actually did not. :)
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:41
@ReneSchipperus: Ohh, then its okay, my bad :(. But seeing how trivial it is, maybe he actually did not. :)
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:41
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
$$|u_n|=frac 1n^frac 23 (1+frac (-1)^nsqrt n)^frac 23$$
$$sim frac 1n^frac 23 $$
thus it is not absolutely convergent.
$$u_n=$$
$$frac (-1)^n-1n^frac 23(1-frac 23frac (-1)^nsqrt n(1+epsilon (n)) )$$
the first series is alternate and the second is absolutely convergent thus $sum u_n $ is conditionally convergent.
answered Dec 20 '17 at 19:37
Salahamam_ Fatima
33.8k21230
It diverges.$,,$
– zhw.
Dec 20 '17 at 19:37
@zhw. yes i edited.
– Salahamam_ Fatima
Dec 20 '17 at 19:38
add a comment |Â
up vote
-1
down vote
"Amusing" downvote to an otherwise perfectly valid answer.
Consider the function $u(x)=x^a$, then $u'(x)=ax^a-1$ hence, if $x_nsim nsim y_n$, $$frac1u(x_n)-frac1u(y_n)=fracu(y_n)-u(x_n)u(x_n)u(y_n)sim(y_n-x_n)fracu'(n)u(n)^2=(y_n-x_n)fracan^a+1$$ The series in the question corresponds to $x_n=n-sqrt n$, $y_n=n+1+sqrtn+1$ and $a=frac23$, then $y_n-x_nsim2sqrt n$ hence $$frac1u(x_n)-frac1u(y_n)simfrac4/3n^2/3+1/2$$ and $frac23+frac12>1$ hence the series converges.
answered Dec 20 '17 at 19:50
Did
243k23208443
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$|u_n|=frac 1n^frac 23 (1+frac (-1)^nsqrt n)^frac 23$$
$$sim frac 1n^frac 23 $$
thus it is not absolutely convergent.
$$u_n=$$
$$frac (-1)^n-1n^frac 23(1-frac 23frac (-1)^nsqrt n(1+epsilon (n)) )$$
the first series is alternate and the second is absolutely convergent thus $sum u_n $ is conditionally convergent.
answered Dec 20 '17 at 19:37
Salahamam_ Fatima
33.8k21230
It diverges.$,,$
– zhw.
Dec 20 '17 at 19:37
@zhw. yes i edited.
– Salahamam_ Fatima
Dec 20 '17 at 19:38
add a comment |Â
up vote
2
down vote
accepted
$$|u_n|=frac 1n^frac 23 (1+frac (-1)^nsqrt n)^frac 23$$
$$sim frac 1n^frac 23 $$
thus it is not absolutely convergent.
$$u_n=$$
$$frac (-1)^n-1n^frac 23(1-frac 23frac (-1)^nsqrt n(1+epsilon (n)) )$$
the first series is alternate and the second is absolutely convergent thus $sum u_n $ is conditionally convergent.
answered Dec 20 '17 at 19:37
Salahamam_ Fatima
33.8k21230
It diverges.$,,$
– zhw.
Dec 20 '17 at 19:37
@zhw. yes i edited.
– Salahamam_ Fatima
Dec 20 '17 at 19:38
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$|u_n|=frac 1n^frac 23 (1+frac (-1)^nsqrt n)^frac 23$$
$$sim frac 1n^frac 23 $$
thus it is not absolutely convergent.
$$u_n=$$
$$frac (-1)^n-1n^frac 23(1-frac 23frac (-1)^nsqrt n(1+epsilon (n)) )$$
the first series is alternate and the second is absolutely convergent thus $sum u_n $ is conditionally convergent.
answered Dec 20 '17 at 19:37
Salahamam_ Fatima
33.8k21230
$$|u_n|=frac 1n^frac 23 (1+frac (-1)^nsqrt n)^frac 23$$
$$sim frac 1n^frac 23 $$
thus it is not absolutely convergent.
$$u_n=$$
$$frac (-1)^n-1n^frac 23(1-frac 23frac (-1)^nsqrt n(1+epsilon (n)) )$$
the first series is alternate and the second is absolutely convergent thus $sum u_n $ is conditionally convergent.
answered Dec 20 '17 at 19:37
Salahamam_ Fatima
33.8k21230
edited Dec 20 '17 at 19:43
answered Dec 20 '17 at 19:37
Salahamam_ Fatima
33.8k21230
answered Dec 20 '17 at 19:37
Salahamam_ Fatima
33.8k21230
answered Dec 20 '17 at 19:37
Salahamam_ Fatima
33.8k21230
33.8k21230
It diverges.$,,$
– zhw.
Dec 20 '17 at 19:37
@zhw. yes i edited.
– Salahamam_ Fatima
Dec 20 '17 at 19:38
add a comment |Â
It diverges.$,,$
– zhw.
Dec 20 '17 at 19:37
@zhw. yes i edited.
– Salahamam_ Fatima
Dec 20 '17 at 19:38
It diverges.$,,$
– zhw.
Dec 20 '17 at 19:37
It diverges.$,,$
– zhw.
Dec 20 '17 at 19:37
@zhw. yes i edited.
– Salahamam_ Fatima
Dec 20 '17 at 19:38
@zhw. yes i edited.
– Salahamam_ Fatima
Dec 20 '17 at 19:38
add a comment |Â
up vote
-1
down vote
"Amusing" downvote to an otherwise perfectly valid answer.
Consider the function $u(x)=x^a$, then $u'(x)=ax^a-1$ hence, if $x_nsim nsim y_n$, $$frac1u(x_n)-frac1u(y_n)=fracu(y_n)-u(x_n)u(x_n)u(y_n)sim(y_n-x_n)fracu'(n)u(n)^2=(y_n-x_n)fracan^a+1$$ The series in the question corresponds to $x_n=n-sqrt n$, $y_n=n+1+sqrtn+1$ and $a=frac23$, then $y_n-x_nsim2sqrt n$ hence $$frac1u(x_n)-frac1u(y_n)simfrac4/3n^2/3+1/2$$ and $frac23+frac12>1$ hence the series converges.
answered Dec 20 '17 at 19:50
Did
243k23208443
add a comment |Â
up vote
-1
down vote
"Amusing" downvote to an otherwise perfectly valid answer.
Consider the function $u(x)=x^a$, then $u'(x)=ax^a-1$ hence, if $x_nsim nsim y_n$, $$frac1u(x_n)-frac1u(y_n)=fracu(y_n)-u(x_n)u(x_n)u(y_n)sim(y_n-x_n)fracu'(n)u(n)^2=(y_n-x_n)fracan^a+1$$ The series in the question corresponds to $x_n=n-sqrt n$, $y_n=n+1+sqrtn+1$ and $a=frac23$, then $y_n-x_nsim2sqrt n$ hence $$frac1u(x_n)-frac1u(y_n)simfrac4/3n^2/3+1/2$$ and $frac23+frac12>1$ hence the series converges.
answered Dec 20 '17 at 19:50
Did
243k23208443
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
"Amusing" downvote to an otherwise perfectly valid answer.
Consider the function $u(x)=x^a$, then $u'(x)=ax^a-1$ hence, if $x_nsim nsim y_n$, $$frac1u(x_n)-frac1u(y_n)=fracu(y_n)-u(x_n)u(x_n)u(y_n)sim(y_n-x_n)fracu'(n)u(n)^2=(y_n-x_n)fracan^a+1$$ The series in the question corresponds to $x_n=n-sqrt n$, $y_n=n+1+sqrtn+1$ and $a=frac23$, then $y_n-x_nsim2sqrt n$ hence $$frac1u(x_n)-frac1u(y_n)simfrac4/3n^2/3+1/2$$ and $frac23+frac12>1$ hence the series converges.
answered Dec 20 '17 at 19:50
Did
243k23208443
"Amusing" downvote to an otherwise perfectly valid answer.
Consider the function $u(x)=x^a$, then $u'(x)=ax^a-1$ hence, if $x_nsim nsim y_n$, $$frac1u(x_n)-frac1u(y_n)=fracu(y_n)-u(x_n)u(x_n)u(y_n)sim(y_n-x_n)fracu'(n)u(n)^2=(y_n-x_n)fracan^a+1$$ The series in the question corresponds to $x_n=n-sqrt n$, $y_n=n+1+sqrtn+1$ and $a=frac23$, then $y_n-x_nsim2sqrt n$ hence $$frac1u(x_n)-frac1u(y_n)simfrac4/3n^2/3+1/2$$ and $frac23+frac12>1$ hence the series converges.
answered Dec 20 '17 at 19:50
Did
243k23208443
edited Aug 25 at 7:23
answered Dec 20 '17 at 19:50
Did
243k23208443
answered Dec 20 '17 at 19:50
Did
243k23208443
answered Dec 20 '17 at 19:50
Did
243k23208443
243k23208443
add a comment |Â
add a comment |Â
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Why not just do a limit comparison with $sum frac1{n^frac23 $?
– Rene Schipperus
Dec 20 '17 at 19:35
Absolute divergence is direct. Are you not asked for (non absolute) convergence?
– Did
Dec 20 '17 at 19:35
@ReneSchipperus, Limit Comparison requires nonnegativity of both the series. The OP's series is not nonnegative always.
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:38
@MrigankaBasuRoyChowdhury I thought he asked about absolute convergence ?
– Rene Schipperus
Dec 20 '17 at 19:40
@ReneSchipperus: Ohh, then its okay, my bad :(. But seeing how trivial it is, maybe he actually did not. :)
– Mriganka Basu Roy Chowdhury
Dec 20 '17 at 19:41