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I am a calculus TA a bit stumped at a question a student has posed me.
What is the method for finding the sum of the series $$\sum_{n=0}^\infty\frac{1+n}{3^n}$$ If you split the numerator up, one becomes a geometric series summing to 3/2, and the other sums to 3/4 by Wolfram Alpha, but I can't explain why.
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:
$$ b^{-1} S = \sum_{k=2}^{\infty} (k-1) b^{-k} $$
$$ S - b^{-1} S = S (1 - b^{-1}) = \sum_{k=1}^{\infty} b^{-k} $$
Consider for $|x| < 1$
$$\sum_{n=1}^\infty x^n = \frac{x}{1-x} .$$
Differentiate and set $x=1/3.$
This is just a different take on (much) earlier answers, mostly to emphasize what can sometimes be gained by playing with the index of summation.
Shifting the index of summation and noting that including $nx^{n-1}$ for $n=0$ adds nothing to the sum, one obtains (for $|x|\lt1$)
$$\sum_{n=0}^\infty(n+1)x^n=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty nx^{n-1}=\left(\sum_{n=0}^\infty x^n\right)'=\left(1\over1-x\right)'={1\over(1-x)^2}$$
Setting $x={1\over3}$ gives
$$\sum_{n=0}^\infty{n+1\over3^n}={9\over4}$$
