Finding values of x for which series converges
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Consider $a_n = (-1)^n frac{x^2+n}{n^2}$ for $n in mathbb{N}$.
For which values of $x in mathbb{R}$ does this series converge?
I applies ratio test and get 1, which does not imply anything clearly. I am not sure if root test does any help or not.
real-analysis sequences-and-series
asked Nov 18 at 14:48
ChakSayantan
986
add a comment |
up vote
0
down vote
favorite
Consider $a_n = (-1)^n frac{x^2+n}{n^2}$ for $n in mathbb{N}$.
For which values of $x in mathbb{R}$ does this series converge?
I applies ratio test and get 1, which does not imply anything clearly. I am not sure if root test does any help or not.
real-analysis sequences-and-series
asked Nov 18 at 14:48
ChakSayantan
986
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider $a_n = (-1)^n frac{x^2+n}{n^2}$ for $n in mathbb{N}$.
For which values of $x in mathbb{R}$ does this series converge?
I applies ratio test and get 1, which does not imply anything clearly. I am not sure if root test does any help or not.
real-analysis sequences-and-series
asked Nov 18 at 14:48
ChakSayantan
986
Consider $a_n = (-1)^n frac{x^2+n}{n^2}$ for $n in mathbb{N}$.
For which values of $x in mathbb{R}$ does this series converge?
I applies ratio test and get 1, which does not imply anything clearly. I am not sure if root test does any help or not.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Nov 18 at 14:48
ChakSayantan
986
asked Nov 18 at 14:48
ChakSayantan
986
asked Nov 18 at 14:48
ChakSayantan
986
asked Nov 18 at 14:48
ChakSayantan
986
asked Nov 18 at 14:48
ChakSayantan
986
986
add a comment |
add a comment |
2 Answers
2
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oldest
votes
up vote
0
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accepted
This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.
$$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$
Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.
answered Nov 18 at 15:48
Mason
1,8251425
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The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$
edited Nov 18 at 15:36
Mason
1,8251425
answered Nov 18 at 14:59
Mohammad Riazi-Kermani
40.3k41958
"all the conditions" meaning $a_nto 0$ as $nto infty$
– Mason
Nov 18 at 15:26
The absolute value of term is decreasing and converges to zero.
– Mohammad Riazi-Kermani
Nov 18 at 15:40
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.
$$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$
Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.
answered Nov 18 at 15:48
Mason
1,8251425
add a comment |
up vote
0
down vote
accepted
This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.
$$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$
Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.
answered Nov 18 at 15:48
Mason
1,8251425
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.
$$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$
Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.
answered Nov 18 at 15:48
Mason
1,8251425
This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.
$$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$
Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.
answered Nov 18 at 15:48
Mason
1,8251425
answered Nov 18 at 15:48
Mason
1,8251425
answered Nov 18 at 15:48
Mason
1,8251425
answered Nov 18 at 15:48
Mason
1,8251425
1,8251425
add a comment |
add a comment |
up vote
0
down vote
The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$
edited Nov 18 at 15:36
Mason
1,8251425
answered Nov 18 at 14:59
Mohammad Riazi-Kermani
40.3k41958
"all the conditions" meaning $a_nto 0$ as $nto infty$
– Mason
Nov 18 at 15:26
The absolute value of term is decreasing and converges to zero.
– Mohammad Riazi-Kermani
Nov 18 at 15:40
add a comment |
up vote
0
down vote
The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$
edited Nov 18 at 15:36
Mason
1,8251425
answered Nov 18 at 14:59
Mohammad Riazi-Kermani
40.3k41958
"all the conditions" meaning $a_nto 0$ as $nto infty$
– Mason
Nov 18 at 15:26
The absolute value of term is decreasing and converges to zero.
– Mohammad Riazi-Kermani
Nov 18 at 15:40
add a comment |
up vote
0
down vote
up vote
0
down vote
The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$
edited Nov 18 at 15:36
Mason
1,8251425
answered Nov 18 at 14:59
Mohammad Riazi-Kermani
40.3k41958
The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$
edited Nov 18 at 15:36
Mason
1,8251425
answered Nov 18 at 14:59
Mohammad Riazi-Kermani
40.3k41958
edited Nov 18 at 15:36
Mason
1,8251425
edited Nov 18 at 15:36
Mason
1,8251425
edited Nov 18 at 15:36
Mason
1,8251425
1,8251425
answered Nov 18 at 14:59
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 14:59
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 14:59
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
"all the conditions" meaning $a_nto 0$ as $nto infty$
– Mason
Nov 18 at 15:26
The absolute value of term is decreasing and converges to zero.
– Mohammad Riazi-Kermani
Nov 18 at 15:40
add a comment |
"all the conditions" meaning $a_nto 0$ as $nto infty$
– Mason
Nov 18 at 15:26
The absolute value of term is decreasing and converges to zero.
– Mohammad Riazi-Kermani
Nov 18 at 15:40
"all the conditions" meaning $a_nto 0$ as $nto infty$
– Mason
Nov 18 at 15:26
"all the conditions" meaning $a_nto 0$ as $nto infty$
– Mason
Nov 18 at 15:26
The absolute value of term is decreasing and converges to zero.
– Mohammad Riazi-Kermani
Nov 18 at 15:40
The absolute value of term is decreasing and converges to zero.
– Mohammad Riazi-Kermani
Nov 18 at 15:40
add a comment |
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