Could anyone help me out with this equation please?
$\displaystyle \sum_{n=2}^{\infty} \frac{2}{(n^{3} - n)3^{n}} = - \frac{1}{2} + \frac{4}{3} \sum_{n=1}^{\infty} \frac {1}{n3^{n}}$
How to prove it? Any ideas?
You have a summation that starts at n=2.
Rewrite the summation of the second term by replacing n-1 by for instance k.
And rewrite the summation of the third term by replacing n+1 by m.
Afterward add or subtract whatever you need to, to make the summations start at 1.
Then you can combine them again.
You can split the summation into 3 summations.
The second summation is:
$\displaystyle \sum_{n=2}^\infty \frac 1 {(n-1)3^n}$
Substitute n=k+1 and you'll get:
$\displaystyle \sum_{k+1=2}^\infty \frac 1 {(k+1-1)3^{k+1}} = \sum_{k=1}^\infty \frac 1 {k 3^{k+1}} = \frac 1 3 \sum_{n=1}^\infty \frac 1 {n 3^n}$
So, I got $\displaystyle \sum_{m=1}^{\infty} \frac{3}{m3^{m}} - \frac{7}{6} + \frac{1}{3}\sum_{k=1}^{\infty} \frac{1}{k3^{k}} - \sum_{n=1}^{\infty} \frac{2}{n3^{n}} $ How to combine them all? I don't know how to cope with those $\displaystyle 2,3, \frac{1}{3} $. For instance, if I go back to n, I get n=0 in the first term, which is not what I want...