convergence of $\sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$

You can prove that it diverges by observing that the sum of each "block" is greater than a constant, that is, $$\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}>C$$for some positive $C$.


Sorry, I have forgotten how to type in LateX, so here it is by hand.

Basically, this is an explicit version of the other answer.

Essentially, we show that each block has a minimal magnitude (infimum) by which it add or subtracts from the partial sum. This non-decreasing (constant) bound makes the series fail the criteria for the alternating series test (I believe)

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Interestingly, if we study the blocks of partial terms, they tend towards "$2$" in magnitude, so the series diverges more "hectically" as we take $n$ to infinity than our bound may suggest.

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Sequences And Series

Real Analysis

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