Normally when I call something a tasty puzzle, it’s a lame local-paper pun about it being to do with cakes or something. In this case, it’s not even that. Sorry to disappoint.

Instead, it’s a puzzle that came to me via reddit:

Find $\sum_{i=1}^{10} \frac{2}{4^{\frac{i}{11}}+2}$.

Eleventh roots? That’s likely to be a brute. Have a go if you want to; spoilers are below the line.

### Higher and lower

A Twitter conversation recently asked how teachers can encourage students to check their answers are sensible, so that they’re not saying things like “this histogram bar ought to be 102cm tall”. My best response to that is the trio of questions: “What’s the biggest it can be?”, “What’s the smallest?”, and “what’s my best guess based on that?”- or rather, can I find sensible bounds on the numbers and use those to inform an estimate?

Looking at the fraction we’re adding up, the value of $4^{\frac{i}{11}}$ is definitely between 1 and 4 over the region we’re looking at - which makes $\frac{2}{4^{\frac{i}{11}}+2}$ somewhere between $\frac{2}{3}$ and $\frac{2}{6}$; the sum is definitely between $\frac{20}{3}$ and $\frac{10}{3}$. A decent guess would be the average of those two, which works out to be 5.

### Doing it properly

I had a few false starts with this – trying to add up all of the terms left me with a mess; trying to use a conjugate trick to get rid of the roots on the bottom even more so. It wasn’t until I hit the idea of dividing everything by 2 that I took off. If we express $4^{\frac{i}{11}}$ as $2^{\frac{2i}{11}}$, we can rewrite the sum as:

$\sum_{i=1}^{10} \frac{1}{2^{\frac{2i-11}{11}}+1}$.

At first glance, that doesn’t look any better. However, if you look at the $i=1$ and $i=10$ terms, they look rather similar:

$i=1$: $\frac{1}{2^{\frac{-9}{11}}+1}$

$i=10$: $\frac{1}{2^{\frac{9}{11}}+1}$

In fact, if you multiply the $i=1$ term top-and-bottom by $2^{\frac{9}{11}}$, it becomes $\frac{2^{\frac{9}{11}}}{1+2^{\frac{9}{11}}}$.

That means adding the first term to the tenth gives you $\frac{1+ 2^{\frac{9}{11}}}{1+2^{\frac{9}{11}}} = 1$.

A similar pattern holds for the 2nd and 9th terms, and for the 3rd and 8th, and so on; there are five such pairs, so the sum works out to be exactly 5.

Which is, of course, in perfect agreement with our earlier guess.