This is part of a bigger exercise.
I am stuck on evaluating. $\sum_{k=2}^\infty \frac{k}{3^k}$
Though have done all the rest of the task..
If you calculate it on wolfram you will find out that: $\sum_{k=1}^\infty \frac{k}{3^k} = \frac{3}{4}$ But why? I need to prove it by hand. I thought about writing it as a cauchy product but i think it is impossible.
Start with $$\sum_{n=0}^\infty z^n = \frac 1 {1-z} \ \ \ \ (\ |z| < 1)$$
Differentiate with respect to $z$ both terms, multiply by $z$, and set $z=\frac 13$. You'll find the solution ;-)
P.S. This of course result from the fact that a power series it is differentiable term by term inside its radius of convergence, and the resulting series has the same radius of convergence
Let $$a_n = \sum_{k=0}^n \frac k {3^k} = \sum_{k=1}^n \frac k {3^k} = \sum_{k=0}^{n-1} \frac {k+1}{3^{k+1}} = \frac 1 3 \sum_{k = 0}^{n-1}\frac {k+1}{3^k}$$ From this we get $$3a_{n+1} - a_n = \sum_{k=0}^n \frac 1 {3^k} \tag{1}$$ From a simple comparison test we know that $\lim_{n\to\infty} a_n$ converges. Denote by $A$ its limit. Taking the limit $n\to\infty$ in (1), we get: $$2A = \sum_{k=0}^{\infty} \frac 1 {3^k} = \frac 3 2$$ This gives the result $A = \frac 3 4$.
