Dear Uncle Colin

I’ve been asked to find the sum of the infinite series $\frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \dots + \frac{2n+1}{2^n} + \dots$. How would you go about it?

Some Expertise Required In Evaluating Sum

Hi, SERIES, and thanks for your message!

I can see a couple of ways of going about it: a sort of “split and recognise” approach, and a more general generating functions approach. There may be others I’ve missed!

### Split and recognise

We can split this up into two series:

• $S_1 = 2\br{\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \dots + \frac{n}{2^n} + \dots}$ and
• $S_2 = \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} + \dots$

$S_2$ is, of course, a well-known geometric series that sums to 1. What about $S_1$?

I’m going to multiply the 2 in first: $S_1 = 1 + \frac{2}{2} + \frac{3}{4} + \dots + \frac{n}{2^{n-1}} + \dots$.

Now, if I let $x=\frac{1}{2}$, that’s the same as $S_1 = 1 + 2x + 3x^2 + \dots + nx^{n-1}$ – which is the binomial expansion of $(1 - x)^{-2}$.

So, with $x= \frac{1}{2}$, that gives $S_1=\br{\frac{1}{2}}^{-2} = 4$.

The series sums to $S_1 + S_2 = 5$. A nice answer!

### Generating functions

Ah, my favourite tool. I can hardly see a series question without wondering if there’s a way to use GFs. This one’s no different.

I’m going to start by letting $x=\frac{1}{2}$ and write the sum as a function of $x$ so I get:

• $G(x) = 3x + 5x^2 + 7x^3 + \dots + (2n+1)x^n + \dots$ (and I’m interested, in the end, in $G\br{\frac{1}{2}}$)

Now, there’s a common difference in the coefficients, which suggests multiplying by $x$ (to shift the coefficients across) and subtracting (to get a simpler polynomial).

• $xG(x) = 3x^2 + 5x^3 + 7x^4 + \dots + (2n+1)x^{n+1} + \dots$
• $(1-x)G(x) = 3x + 2\br{x^2 + x^3 + x^4 + \dots}$

That thing in the big bracket is a geometric series with first term $x^2$ and common ratio $x$:

• $(1-x)G(x) = 3x + 2\frac{x^2}{1-x}$

And now we can just substitute in $x= \frac{1}{2}$! (I was tempted to divide the bracket over, but it’s probably simpler not to here.)

• $\frac{1}{2}G(x) = \frac{3}{2} + 2\frac{1/4}{1/2}$
• That final fraction needs a bit of care: multiplying the 2 in means doubling the top, and makes it $\frac{1/2}{1/2}=1$.
• $\frac{1}{2}G(x) = \frac{3}{2} + 1$
• Double it all: $G(x) = 3 + 2 = 5$, as before!

I’m not saying the second way is easier, just that it’s interesting, and the kind of approach you can apply elsewhere!

Hope that helps,

- Uncle Colin

• Edited 2022-07-01 to fix a LaTeX error. Thanks to Adam for pointing it out.