<s>[INST] Why the following series doesn't absolute convergence $\sum_{k=1}^{\infty} (-1)^k(\sqrt{k^2+1}-\sqrt{k^2-1})$ I have the following series, I don't understand why it doesn't converge absolutely. 
$$\sum_{k=1}^{\infty} (-1)^k(\sqrt{k^2+1}-\sqrt{k^2-1})$$
This is what I did:
$$\sum_{k=1}^{\infty} |(-1)^k(\sqrt{k^2+1}-\sqrt{k^2-1})|=\sum_{k=1}^{\infty} |(\sqrt{k^2+1}-\sqrt{k^2-1})*\frac{\sqrt{k^2+1}+\sqrt{k^2-1}}{\sqrt{k^2+1}+\sqrt{k^2-1}}|=\sum_{k=1}^{\infty}\frac{2}{\sqrt{k^2+1}+\sqrt{k^2-1}} \geq \sum_{k=1}^{\infty} \frac{1}{\sqrt{k^2+1}} \geq\sum_{k=1}^{\infty} \frac{1}{2k^2}$$
And $\sum_{k=1}^{\infty} \frac{1}{2k^2}$ convergences therefore $\sum_{k=1}^{\infty} (-1)^k(\sqrt{k^2+1}-\sqrt{k^2-1})$ is absolutely convergent.
I don't understand why it's not true.
Any ideas? Thank you!
 [/INST] There are three points I want to make.
Firstly, your inequality is incorrect. Let me show you the correct one:
\begin{align*}
\sum_{k=1}^n(\sqrt{k^2+1}-\sqrt{k^2-1})={}&\sum_{k=1}^n(\sqrt{k^2+1}-\sqrt{k^2-1})\frac{\sqrt{k^2+1}+\sqrt{k^2-1}}{\sqrt{k^2+1}+\sqrt{k^2-1}}={} \\
{}={}&\sum_{k=1}^n\frac{2}{\sqrt{k^2+1}+\sqrt{k^2-1}}\leq\sum_{k=1}^n\frac{2}{\sqrt{k^2+1}},
\end{align*}
because when you reduce a denominator, the fraction increases.
Secondly, the above majoring series doesn't converge anyway, because it is asymptotic to $\sum\frac{1}{k}$, the harmonic series, which is well-known not to converge.
Thirdly, the starting series does not converge, because:
\begin{align*}
\frac{2}{\sqrt{k^2+1}+\sqrt{k^2-1}}={}&\frac{2}{k}\frac{1}{\sqrt{1+\frac{1}{k^2}}+\sqrt{1-\frac{1}{k^2}}}={} \\
{}={}&\frac{2}{k}\frac{1}{1+\frac{2}{k^2}+o\left(\frac{1}{k^2}\right)+1-\frac{2}{k^2}+o\left(\frac{1}{k^2}\right)}={} \\
{}={}&\frac{2}{2k}\frac{1}{1+o\left(\frac{1}{k^2}\right)}\sim\frac{1}{k},
\end{align*}
the Harmonic series once more.
Note: I assumed you knew that $x_n\sim y_n$ for $n\to\infty$ implies the convergence of $\sum x_n$ and $\sum y_n$ is the same. This is proved by noting asymptotic means there are constants $a,b$ such that eventually $ax_n\leq y_n\leq bx_n$, so if $\sum x_n$ converges so does $\sum bx_n$ and thus so does $\sum y_n$, whereas if $\sum x_n$ doesn't converge then neither does $\sum ax_n$ and thus neither does $\sum y_n$. That works at least with positive terms as in the case where I used it. With alternating signs there might be extra cancellations made possible by a majoring.
</s>