Show that $\sum_{k=1}^\infty \frac{i}{k(k+1)}$ converges. Find its sum.
The presence of the $i$ throws me off. Our professor taught us to find $a$ (the first term in the sequence) and $z$ (the multiplier to get the next term). Then find the modulus of z, and if $|z|<1$ then the series converges.
Here $a=\frac{i}{2}$, but I'm struggling to find $z$. I think that $z=i-\frac{i}{K+1}$, but I'm not completely confident in that. Additionally, if that $z$ value is correct would I end up with $\sqrt{0^2+(1-\frac{1}{K+1})^2}$? I'm pretty sure there shouldn't be a variable in there...
Also how should I find the sum?
As suggested by @Kavi Rama Murthy, pull out the constant multiple of $i$ and simply do partial fractions:
$${=i\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)}$$
So you only need to find the answer to the infinite sum in the brackets, and multiply the result by $i$. Can you take it from here?