13
$\begingroup$

I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: $$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}$$

my steps:

$$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\sum_{n=1}^{\infty}\frac{2}{4n^2-1}-\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=..help..=sum$$

I am lacking some important properties, I feel I am coming to the right step and cannot spit that out..

$\endgroup$
  • 1
    $\begingroup$ Hint: Write your rational functions as a linear combination of fractions with denominator being a linear function. $\endgroup$ – Ofir Dec 26 '12 at 10:47
  • 1
    $\begingroup$ see math.stackexchange.com/questions/238728/finding-summation/… $\endgroup$ – MathOverview Dec 26 '12 at 10:48
  • 1
    $\begingroup$ @doniyor taking $\displaystyle \lim_{n\to \infty} \left(\frac{1}{2} - \frac{1}{4n+2}\right)$ in the result posted in Elias's link you get $\frac{1}{2}$ $\endgroup$ – Rustyn Dec 26 '12 at 10:55
6
$\begingroup$

Hint: Partial Fraction decomposition:$$\frac{1}{4n^2-1}=\frac{1}{(2n-1)(2n+1)}=\frac12[\frac{1}{2n-1}-\frac{1}{2n+1}]$$ You must then compute the closed form of $$\sum_{n=1}^k[\frac{1}{2n-1}-\frac{1}{2n+1}]$$ Can you do that? Note that $$\sum_{n=1}^k\frac{1}{2n-1}=\frac11+\frac13+...+\frac1{2k-1}=\frac1{2\cdot 0+1}+\frac1{2\cdot 1+1}+...+\frac1{2(k-1)+1}=\sum_{n=0}^{k-1}\frac{1}{2n+1}=\sum_{n=1}^{k}\frac1{2n+1}+\frac{1}{2\cdot 0+1}-\frac1{2k+1}$$

$\endgroup$
  • $\begingroup$ can you pls expand this step? $\sum_{n=1}^k\frac{1}{2n-1}=\sum_{n=0}^{k-1}\frac{1}{2n+1}$ $\endgroup$ – doniyor Dec 26 '12 at 11:29
  • $\begingroup$ @doniyor Sure I can. $\endgroup$ – Nameless Dec 26 '12 at 11:29
16
$\begingroup$

Note $\frac{1}{4n^2-1}=\frac{1}{(2n+1)(2n-1)}={\frac{1}{2}}\times\frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)}={\frac{1}{2}}\times[\frac{1}{2n-1}-\frac{1}{2n+1}]$ for $n\in\mathbb N.$

Let for $k\in\mathbb N,$ $S_k=\displaystyle\sum_{n=1}^{k}\frac{1}{4n^2-1}$ $\implies S_k={\frac{1}{2}}\displaystyle\sum_{n=1}^{k}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right].$ Thus for $k=1,2,...$

$S_1={\frac{1}{2}}\displaystyle\sum_{n=1}^{1}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]=\frac{1}{2}(1-\frac{1}{3})$

$S_2={\frac{1}{2}}\displaystyle\sum_{n=1}^{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})]=\frac{1}{2}(1-\frac{1}{5})$

$S_3={\frac{1}{2}}\displaystyle\sum_{n=1}^{3}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})]=\frac{1}{2}(1-\frac{1}{7})$

...

$S_k=\frac{1}{2}(1-\frac{1}{2k+1})$

$\implies\displaystyle\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\displaystyle\lim_{k\to\infty}S_k=\frac{1}{2}.$

$\endgroup$
  • $\begingroup$ great, thanks Sugata $\endgroup$ – doniyor Dec 26 '12 at 13:39
  • $\begingroup$ You're most welcome. $\endgroup$ – Sugata Adhya Dec 26 '12 at 13:41
6
$\begingroup$

Or we want to compute it fastly and use the formula $$\sum_{k=1}^\infty \frac1{k^2-x^2}=\frac1{2x^2}-\frac{\pi\cot\,\pi x}{2x}$$ where $x=\frac{1}{2}$ because $$\sum_{k=1}^{\infty}\frac{1}{4k^2-1}=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2-\left(\frac{1}{2}\right)^2}$$ Here you may find more information about this precious way.

$\endgroup$
  • $\begingroup$ @doniyor: you're welcome! $\endgroup$ – user 1357113 Dec 26 '12 at 20:52
3
$\begingroup$

Hint: Work on $S_n=\sum_{k=1}^n\frac{1}{4k^2-1}$ and take its limit when $n\to\infty$. Note that $$\frac{1}{4n^2-1}=\frac{1}{2(2n-1)}-\frac{1}{2(2n+1)}$$

$\endgroup$
  • 1
    $\begingroup$ Why don't you experts leave such problems for beginners like us ? :( $\endgroup$ – Sugata Adhya Dec 26 '12 at 13:40
  • $\begingroup$ @SugataAdhya: Dear Sugata, I am not an expert. I am a teacher and like to help others in Maths. That's it. We are here to help eachother in learning Maths well. :-) $\endgroup$ – mrs Dec 26 '12 at 16:13
  • $\begingroup$ :), yeah, but you guys are doing wonderful job here.. thanks again for all you $\endgroup$ – doniyor Dec 26 '12 at 20:50
  • $\begingroup$ Yes, as dear doniyor says, you're doing a wonderful job here! + $\endgroup$ – Namaste Mar 2 '13 at 2:45
0
$\begingroup$

This is an easy problem by using Fourier's serie of $|\sin(x)|$. So, $|\sin(x)|=\dfrac2\pi-\dfrac4\pi\sum_{n=1}^{\infty}\dfrac{\cos(2nx)}{4n^2-1} $. By taking $x=0$, we obtain:

$0=\dfrac2\pi-\dfrac4\pi\sum_{n=1}^{\infty}\dfrac{1}{4n^2-1} $.

So,

$\sum_{n=1}^{\infty}\dfrac{1}{4n^2-1}=\dfrac12$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.