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I was thinking about this problem:

Given a series $\sum_{n=4}^{\infty}\frac{1}{n^{2}-1}$ ,how can i show that its sum is a/an rational/irrational number,given that the series converges?

Could someone point me in the right direction? Thanks everyone in advance.

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6  
Partial fractions. –  Gerry Myerson Dec 2 '12 at 6:21

2 Answers 2

up vote 5 down vote accepted

$1/(n-1)-1/(n+1) = 2/(n^2-1) $, so $$\sum_{n=4}^{\infty} \frac{1}{n^2-1} = (1/2)\sum_{n=4}^{\infty} \left(\frac{1}{n-1} - \frac{1}{n+1}\right) = \frac{1}{3} + \frac{1}{4} $$ since all the later terms are cancelled out.

Whoops - as pointed out by Limitless, this should be $\frac{1}{2}\left(\frac{1}{3} + \frac{1}{4}\right)$.

Note that this also allows you to get an explicit expression for $\sum_{n=a}^b \frac{1}{n^2-1}$ for any integers $a$ and $b$.

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@martycohen Thank you,sir. I have got it.. –  learner Dec 2 '12 at 7:17
3  
@martycohen, this is incorrect as stated: $$(1/2)\sum_{n=4}^{\infty} \left(\frac{1}{n-1} - \frac{1}{n+1}\right) \ne \frac{1}{3} + \frac{1}{4}.$$ Instead, we have that $$(1/2)\sum_{n=4}^{\infty} \left(\frac{1}{n-1} - \frac{1}{n+1}\right)=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}\right).$$ –  000 Dec 2 '12 at 8:37
    
Thanks for correcting my oversight. I will edit my answer with acknowledgement. –  marty cohen Dec 2 '12 at 23:50

Find the closed form and take the limit in the case of infinite sums. That is, you find the closed form of the sum $\sum_{4 \le k \le m}\frac{1}{k^2-1}$ and evaluate $\lim_{m \to \infty}(\text{closed form of the sum})$.

In this case, you apply partial fraction decomposition to $\frac{1}{k^2-1}$ and arrive at $$\frac{1}{k^2-1}=\frac{-\frac{1}{2}}{k+1}+\frac{\frac{1}{2}}{k-1}=\frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right).$$

Taking the partial sum $$\sum_{4 \le k \le m}\frac{1}{k^2-1}=\frac{1}{2}\sum_{4 \le k \le m}\frac{1}{k-1}-\frac{1}{k+1}.$$

The sum telescopes, so we see that $$\sum_{4 \le k \le m}\frac{1}{k+1}-\frac{1}{k-1}=\color{red}{\frac{1}{3}}-\color{green}{\frac{1}{5}}+\color{red}{\frac{1}{4}}-\frac{1}{6}+\color{green}{\frac{1}{5}}-\frac{1}{7}+\dots+\color{green}{\frac{1}{m-1}}-\frac{1}{m-3}+\color{red}{\frac{1}{m}}-\frac{1}{m-2}+\color{red}{\frac{1}{m+1}}-\color{green}{\frac{1}{m-1}}=\frac{1}{3}+\frac{1}{4}+\frac{1}{m}+\frac{1}{m+1}.$$

Taking the limit, we have $$\lim_{m \to \infty}\frac{1}{3}+\frac{1}{4}+\frac{1}{m}+\frac{1}{m+1}=\frac{1}{3}+\frac{1}{4}+\lim_{m \to \infty}\frac{1}{m}+\frac{1}{m+1}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}.$$

Recalling the factor of $\frac{1}{2}$, we arrive at $\sum_{4 \le k \le \infty}\frac{1}{k^2-1}=\frac{1}{2}\cdot\frac{7}{12}=\frac{7}{24}$.

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thanks a lot.It is crystal clear now. –  learner Dec 2 '12 at 7:42
    
@learner, I'm glad! I love colors! –  000 Dec 2 '12 at 8:36
3  
I love the use of colors! I would recommend something other than red and green though, considering red-green is the most common type of colorblindness. –  Antonio Vargas Dec 2 '12 at 23:58
    
@AntonioVargas, LOL. What luck I have! I will keep that in mind. :) –  000 Dec 3 '12 at 0:51

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