Krdytkyu

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

I've tried to rewrite the questions as $$sum _{n=2}^{infty ::}left(-frac{1}{2left(n+1right)}+frac{1}{2left(n-1right)}right)$$ 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?

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

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

$$sum_{n=2}^{infty}frac{1}{n^2-1}=frac12sum_{n=2}^{infty}left(frac{1}{n-1}-frac{1}{n+1}right)$$
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 $frac{3}{4}$.

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 $$sum_{n=2}^{infty}frac{1}{n-1}=sum_{n=1}^{infty}frac{1}{n} qquad text{and}qquadsum_{n=2}^{infty}frac{1}{n+1}=sum_{n=3}^{infty}frac{1}{n}$$ Hence
$$sum_{n=2}^{infty}frac{1}{n^2-1}=frac12sum_{n=2}^{infty}left(frac{1}{n-1}-frac{1}{n+1}right)=frac12left(sum_{n=1}^{infty}frac1n-sum_{n=3}^{infty}frac1nright)=frac12left(frac11+frac12right)=frac34$$

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}^nfrac1{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}|\
&lefrac1{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
quadtext{ 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:

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