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I'm trying to understand the first proof suggested here http://math.cmu.edu/~bwsulliv/basel-problem.pdf for the basel problem.

The writer writes:

$\frac{\pi \cos(\pi y)}{\sin(\pi y)} = \frac{1}{y} - \sum_{k=1}^{\infty}\frac{2y}{k^2 - y^2}$ which is true and I understand why it is true. Lets call this formula A.

The problem is with the next part. After this line, the author writes:

$\frac{1}{y} + \sum_{k=1}^{\infty}\frac{1}{k^2 - y^2} = \frac{1}{2y^2}-\frac{\pi \cos (\pi y)}{2y\sin(\pi y)}$

I'm not sure why that's true.

Lets work from formula A and divide it by $-2y$:

$-\frac{\pi \cos(\pi y)}{2y\sin(\pi y)} = -\frac{1}{2y^2} + \sum_{k=1}^{\infty}\frac{1}{k^2 - y^2}$

Now lets add $\frac{1}{2y^2}$ to both sides:

$\frac{1}{2y^2} -\frac{\pi \cos(\pi y)}{2y\sin(\pi y)} = \sum_{k=1}^{\infty}\frac{1}{k^2 - y^2}$

Where did the $\frac{1}{y}$ go? Am I missing something or is the author wrong?

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    You are correct: It's a typo. Leave it away. In the next term/line it disappears again ;) – 2017-01-14
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    Lol, typos? Pretty common, we're all human (I'm watching for you trolling non-human account) – 2017-01-14
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    Well if you ask my friends, mathematicians are not human so...You never know... – 2017-01-14

2 Answers 2

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Yes, you are correct that this is an error. The error disappears later in the calculation, and the calculation works (when done correctly)

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I am the author of the presentation that you linked to. You are correct that this is just a typo. (We all make mistakes!) I will update the file that I have accordingly. However, because I am no longer a student at CMU, I lack the login credentials to update the file at that link. I hope you still found the presentation useful. Cheers!