Sum of the series: $\sum_{n=0}^\infty\frac{1+n}{3^n}$

Using the usual geometric series, you have

$$\frac{x}{1-x} = \sum_{n=1}^{\infty} x^n $$ thus differentiating you get

$$\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$$ thus

$$\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^2}$$ Then

$$\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^2}$$

So your sum is just

$$\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{1/3}{(1-1/3)^2} = \frac{3}{4}$$

Michael, there is a general trick here that can be very useful.

Think of $xD$ as an operator on differentiable functions $p(x)$ that "differentiates and then multiplies by $x$". Then polynomials in $xD$ are also operators on differentiable functions, if one interprets powers of $xD$ by composition. For example if $f(x)=2x^{2}+3$ then $f(xD)$ applied to a differentiable function p(x) gives $2xD(xD(p(x)))+3p(x)$. (Think of 3 as the operator "multiplication by 3". )

Now you should verify the following very useful fact:

The operator $p(xD)$ applied term by term to the power series $\sum_{n}a_{n}x^{n}$ gives the power series $\sum_{n}p(n)a_{n}x^{n}$.

In particular, your series has the form $\sum f(n)x^{n}$ evaluated at $x=1/3$, where $f(n)=n+1$. So to compute the sum, apply $1+xD$ to the function $1/(1-x)=1+x+x^{2}+\ldots$ and then evaluate at $x=1/3$.

I think I learned this from Polya and Szego's "Problems and Theorems in Analysis", which is absolutely stuffed with mathematical wisdom.

I believe the other answers about differentiating and such are much cleaner, but I thought I would add that there is an even more elementary way of doing the sum:

$$ S = \sum_{k=1}^{\infty} k b^{-k} $$

by using the standard trick of multiplying by $b^{-1}$ then subtracting:

$$ b^{-1} S = \sum_{k=2}^{\infty} (k-1) b^{-k} $$

$$ S - b^{-1} S = S (1 - b^{-1}) = \sum_{k=1}^{\infty} b^{-k} $$