# Find the sum: $\sum_{n=2}^\infty \frac{1}{n^2-1}$

Evaluate : $$\sum_{n=2}^\infty \frac{1}{n^2-1}$$

I've tried to rewrite the questions as $$\sum _{n=2}^{\infty \:\:}\left(-\frac{1}{2\left(n+1\right)}+\frac{1}{2\left(n-1\right)}\right)$$ but still couldnt't get any answer as when I substitute numbers $$(n)$$ into the $$Sn \left(-\frac{1}{6}+\frac{1}{2}-\frac{1}{8}+\frac{1}{4}-\frac{1}{10}\right)$$ , $$Sn$$ just keeps going on and on. So how do I solve this problem and what does this series converge to?

• What does mean $[.]$ ? Commented Nov 23, 2018 at 3:03
• That was me @Nosrati, its just a square bracket to encase the argument of the sum. Commented Nov 23, 2018 at 3:05
• Keyword: telescoping. Commented Nov 23, 2018 at 3:50
• Is my answer okay? Commented Nov 23, 2018 at 3:51

$$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2(n-1)}-\frac{1}{2(n+1)}$$

Taking $$\frac{1}{2}$$ common ,

$$\sum_{n=2}^{\infty}\frac{1}{n^2-1}=\frac12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$ Write the first few terms of the series as :

$$\frac{1}{2}( 1 - \frac{1}{3} +\frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \cdot \cdot \cdot$$

You can see that except $$1$$ and $$\frac{1}{2}$$ every terms get cancelled out and the $$n$$-th term tends to zero. So the result inside the bracket is $$\frac{3}{2}$$ . But there is a $$\frac{1}{2}$$ outside the bracket which needs to be multiplied. So the answer is $$\frac{3}{4}$$.

• Yes, this looks ok to me. Commented Nov 23, 2018 at 4:27

As you correctly have: $$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2(n-1)}-\frac{1}{2(n+1)}$$ Now, observe that $$\sum_{n=2}^{\infty}\frac{1}{n-1}=\sum_{n=1}^{\infty}\frac{1}{n} \qquad \text{and}\qquad\sum_{n=2}^{\infty}\frac{1}{n+1}=\sum_{n=3}^{\infty}\frac{1}{n}$$ Hence $$\sum_{n=2}^{\infty}\frac{1}{n^2-1}=\frac12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)=\frac12\left(\sum_{n=1}^{\infty}\frac1n-\sum_{n=3}^{\infty}\frac1n\right)=\frac12\left(\frac11+\frac12\right)=\frac34$$

• If you consider partial sums, you can avoid writing out the divergent harmonic series which makes me quite comfortable :) Commented Nov 23, 2018 at 3:35
• @Szeto: +1 to that. I can prove many things if I'm allowed to use expressions like $\sum_{n=1}^{\infty}\frac{1}{n}$.... Commented Nov 23, 2018 at 3:39
• Is my answer okay please check once Commented Nov 23, 2018 at 3:56

More generally, if $$s_m(n) =\sum_{k=m+1}^n \frac{1}{k^2-m^2}$$ then, if $$n > 3m$$,

$$\begin{array}\\ s_m(n) &=\sum_{k=m+1}^n \frac{1}{k^2-m^2}\\ &=\sum_{k=m+1}^n \frac1{2m}(\frac{1}{k-m}-\frac1{k+m})\\ &=\frac1{2m}\sum_{k=m+1}^n \frac{1}{k-m}-\frac1{2m}\sum_{k=m+1}^n\frac1{k+m}\\ &=\frac1{2m}(\sum_{k=1}^{2m} \frac{1}{k}+\sum_{k=2m+1}^{n-m} \frac{1}{k})-(\frac1{2m}\sum_{k=2m+1}^{n-m}\frac1{k}+\frac1{2m}\sum_{k=n-m+1}^{n+m}\frac1{k})\\ &=\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}-\frac1{2m}\sum_{k=n-m+1}^{n+m}\frac1{k}\\ \end{array}$$

so

$$\begin{array}\\ s_m(n)-\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k} &=-\frac1{2m}\sum_{k=n-m+1}^{n+m}\frac1{k}\\ \text{so that}\\ |s_m(n)-\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}| &=\frac1{2m}|\sum_{k=n-m+1}^{n+m}\frac1{k}|\\ &\le\frac1{2m}|\sum_{k=n-m+1}^{n+m}\frac1{n-m+1}|\\ &=\frac1{2m}|\frac{2m}{n-m+1}|\\ &=\frac{1}{n-m+1}\\ &\to 0 \quad\text{ as } n \to \infty\\ \end{array}$$

Therefore $$\lim_{n \to \infty} s_m(n) =\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}$$.

For $$m=1$$ the sum is $$\frac1{2}(\frac1{1}+\frac1{2}) =\frac34$$.

HINT:

$$\frac{1}{2[(n+2)-1]}-\frac{1}{2[n+1]}=0$$

• Is my answer okay? Please check Commented Nov 23, 2018 at 3:56