Evaluating power series

Question

How do I evaluate the power series $$\sum_{n = 1}^{\infty}\dfrac{n}{9^{n}}$$ without using the formula for infinite geometric series?

I am interested in an approach which doesn't make use of infinite geometric series.

Any help would be highly appreciated. New approaches are most welcomed. Thanks.

Answer 1

I'm not sure about the exact value, but you can compare it to the integral, which converges since $|\frac{1}{9}|<1$, $\int_{0}^{\infty}x a^x dx$, which will be of the same order, and then derive the difference using Bernoulli polynomials.

Answer 2

Check with induction that the partial sum is $$\sum_{k=1}^N \frac{k}{9^k} = \frac{9 - 9^{-N}(8N+9)}{64}$$ Evaluate the limit when $N$ goes to infinity and you get the result.

Answer 3

Notice that $x \frac{d}{dx} \frac{1}{1-x} = \sum_{i=0}^{\infty}ix^i $. So differenciating you get $$x \frac{d}{dx} \frac{1}{1-x} = \frac{x}{(1-x)^2}$$ Substituting $x= \frac{1}{9}$ you get required answer.

Answer 4

Denote $A=\sum_{n\geq 1} n9^{-n}$, $B=\sum_{n\geq 2} n9^{-n}$, $C=\sum_{n\geq 1} 9^{-n}$, $D=\sum_{n\geq 2} 9^{-n}$. You have the closed system $$ A=B+\frac 19,\quad 9B-A=C,\quad C=D+\frac 19,\quad 9D=C, $$ which gives the value of $A$. Of course, as a byproduct of the computation one also gets the value of the geometric series $C$.