$$\ \sum_{n=5}^\infty \frac{8}{n^2-1} $$

I tried the following:

$$\ \sum_{n=5}^\infty \frac{8}{n^2-1} = \sum_{n=5}^\infty \frac{8}{n-1} - \frac{8}{n} =$$ $$\left(2-\frac{8}{5}\right) + \left(\frac{8}{5} - \frac{8}{6}\right) + \left(\frac{8}{6} - \frac{8}{7}\right) + \cdots + \left(-\frac{8}{n}\right)$$

Terms cancelled each other out, therefore we are left with: $$ \ (2 - \frac{8}{n}) $$

I could think the series converges to 2 since: $$\lim_{n \to \infty} \left(2-\frac 8 n \right) = 2$$

However, the correct answer is $$ \frac{9}{5} $$ what am I doing wrong?

  • 3
    $\begingroup$ In the second line consider rather $\,\dfrac 8{n^2-1}=\dfrac 4{n-1}-\dfrac 4{n+1}$ $\endgroup$ – Raymond Manzoni Jul 5 '17 at 22:50
  • 3
    $\begingroup$ The first equation in what you tried is not correct. $\left(n-1\right)n \ne n^2 -1$ $\endgroup$ – Andrew Jul 5 '17 at 22:51

Your partial fraction is wrong:


Using similar trick, you will see that most terms cancel out and you will left with $1+\frac45.$


to obtain the partial fractions,

Since $n^2-1=(n-1)(n+1)$,

$$\frac{8}{(n-1)(n+1)}= \frac{A}{n+1}+\frac{B}{n-1}.$$

We can for instance equate the two and solve for $A$ and $B$ by comparing coefficients.

I use a trick call heaviside cover method.

To determinte $A$, $n+1=0$, $n=-1$. I will cover up the term $n+1$ in the denominator of the left hand side and evalute $\frac{8}{n-1}$ with $n=-1$, hence $A=-4$.

We can do similar stuff for $B$ as well.

  • $\begingroup$ Thank you! Any way you can tell me how to get the partial fraction right and how is that equal to -4/n+1 + 4/n-1 $\endgroup$ – chris Jul 5 '17 at 23:12
  • $\begingroup$ edited my post. $\endgroup$ – Siong Thye Goh Jul 5 '17 at 23:19

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