Does $\sum _{n=1}^{\infty }\left(-1\right)^{n+1}\left(1-\cos\left(\frac{1}{\sqrt{n}}\right)\right)$ converges conditionally? I'm trying to understand whether the following series converges absolutely, conditionally or diverges. 
$$
\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\left(1-\cos\left(\frac{1}{\sqrt{n}}\right)\right)
$$
I think that it converges conditionally. But it's getting complicated for me to show that it does not converge absolutely. I've already tried the limit comparison test with $b_n = \frac{1}{\sqrt{n}}$ but the limit results in $0$. I've also tried the integral test, as the conditions for the test hold, but the integral became very complicated. 
Is there an easier way?
 A: Using the alternating series test, we see that $\sum_{n=1}^\infty (-1)^{n+1}\left( 1-\cos\left(\frac{1}{\sqrt{n}}\right)\right)$ converges. 
From Taylor series we have
$$\cos\left(\frac{1}{\sqrt{n}}\right) \leq 1-\frac{\left(\frac{1}{\sqrt{n}}\right)^2}{2} + \frac{\left(\frac{1}{\sqrt{n}}\right)^4}{24}= 1-\frac{1}{2n} + \frac{1}{24n^2}\quad \text{for $n\geq 1$}$$
Therefore, 
$$\sum_{n=1}^\infty \left(1-\cos\left(\frac{1}{\sqrt{n}}\right) \right)\geq \sum_{n=1}^\infty \left(\frac{1}{2n} -\frac{1}{24n^2}\right)= \infty$$
So, the series fails to converge absolutely.
A: Note that 
$$1-\cos\left(\frac{1}{\sqrt{n}}\right)= \frac1{2n}+O\left(\frac1{n\sqrt n}\right)$$
then the given series doesn’t converge absolutely by limit comparison test with $\sum \frac1{n}$ and then
$$
\sum _{n=1}^{\infty }\left(-1\right)^{n+1}\left(1-\cos\left(\frac{1}{\sqrt{n}}\right)\right)=\sum _{n=1}^{\infty }\left(-1\right)^{n+1} \frac1{2n}+\sum _{n=1}^{\infty }\left(-1\right)^{n+1} O\left(\frac1{n\sqrt n}\right)
$$
which converges since


*

*$\sum _{n=1}^{\infty }\left(-1\right)^{n+1} \frac{1}{2n}$ converges by alternating series test

*$\sum _{n=1}^{\infty }\left(-1\right)^{n+1} O\left(\frac1{n\sqrt n}\right)$ converges absolutely by limit comparison test with $\sum \frac1{n\sqrt n}$
A: I think the integral test works.
The taylor series of $\cos(x)$ is
$$\sum^\infty_{k=0}\frac{(-1)^k}{(2k)!}x^{2k}$$
Thus, $$1-\cos(\frac1{\sqrt{n}})= -\sum^\infty_{k=1} \frac{(-1)^k}{(2k)!}\frac1{n^{k}}$$
$$-\int^\infty_1 \sum^\infty_{k=1} \frac{(-1)^k}{(2k)!}\frac1{x^{k}}dx=-\int^\infty_1 \sum^\infty_{k=2} \frac{(-1)^k}{(2k)!}\frac1{x^{k}}dx+\int^\infty_1\frac{1}{2x}dx=C+\infty$$ where $C$ is finite.
Therefore the sum does not converge absolutely.
It is an elementary exercise to prove $C$ is finite.:)
