How to evaluate \sum_{n=1}^{n=\infty}(\sum_{k=n}^{k=n^2}\frac{1}{k^2})

caldaridvq

caldaridvq

Answered question

2022-04-20

How to evaluate n=1n=(k=nk=n21k2)

Answer & Explanation

louran20z47

louran20z47

Beginner2022-04-21Added 14 answers

n=1n=({k=n}k=n21k2)=k=1n=|k|1k2
=k=1k[k]+1k2
k=1kkk2
=k=1(1k1k32
which clearly diverges.

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