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

Solution 1:

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 $C$.

Solution 2:

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)

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.

Solution 3:

This does not converge; it oscillates by approximately $2$.

If $m^2 \le m \le m^2+2n $, then $\lfloor \sqrt{n}\rfloor = m$.

Therefore

$\begin{array}\\ S(M) &=\sum_{n=1}^{M^2+2M} \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}\\ &=\sum_{m=1}^{M} \sum_{n=m^2}^{m^2+2m} \frac{(-1)^m}{\sqrt{n}}\\ &=\sum_{m=1}^{M} (-1)^m\sum_{n=m^2}^{m^2+2m} \frac{1}{\sqrt{n}}\\ &=\sum_{m=1}^{M} (-1)^ms(m) \qquad\text{where }s(m)=\sum_{n=m^2}^{m^2+2m} \frac{1}{\sqrt{n}}\\ \end{array} $

$s(m) \lt \frac{2m+1}{m} = 2+\frac1{m} $ and $s(m) \gt \frac{2m+1}{m+1} =2-\frac1{m+1} $.

Therefore, as $n$ goes from $m^2$ to $m^2+2n$, the value of the sum changes by approximately $2$, increasing when $n$ is even and decreasing when $n$ is odd.

Therefore the series does not converge, since it oscillates.

However, there might be a sense in which the series converges since we can write $S(2M) =\sum_{m=1}^{2M}(-1)^ms(m) =\sum_{m=1}^{M}(s(2m)-s(2m-1)) $.

I can show that that $s(m)-s(m+1) \to 0$, but I can not yet show that $\lim_{M \to \infty}\sum_{m=1}^{M}(s(2m)-s(2m-1)) $ exists.

Here's what I have so far, and I will leave it at this:

$\begin{array}\\ s(m)-s(m+1) &\gt (2-\frac1{m+1})-(2+\frac1{m+1})\\ &= -\frac{2}{m+1}\\ \text{and}\\ s(m)-s(m+1) &\lt (2+\frac1{m})-(2-\frac1{m+2})\\ &= \frac1{m}+\frac1{m+2}\\ &= \frac{2m+2}{m(m+2)}\\ \text{so}\\ s(m)-s(m-1) &= O(\frac1{m})\\ \end{array} $