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!
 A: Based on the direction of the inequality comparing with $\sum\frac1{2k^2}$, you cannot conclude that your series converges. It should be the other way around. Also you should compare with $\frac 1{2k}$ because of the square root, which does not converge.
A: Most of your calculations are correct.
But to show the series is bigger than a convergent series does not show the first series is convergent.
Also, $1/\sqrt{k^2+1}$ is more than $1/(k+1)$, and the sum of $1/(k+1)$ is divergent.  That does show the first series is divergent.
A: There no absolute convergence because 
\begin{align*}\sqrt{k^2+1}-\sqrt{k^2-1}&=k\biggl(\sqrt{1+\frac1{k^2}}-\sqrt{1-\frac1{k^2}}\biggr)\\&=k\biggl(1+\frac1{2k^2}+o\Bigl(\frac1{k^2}\Bigr)-1+\frac1{2k^2}+o\Bigl(\frac1{k^2}\Bigr)\biggr)\\&=\frac1k+o\Bigl(\frac1{k}\Bigr)\color{red}{\sim_\infty\frac1k}\end{align*}
As the harmonic series diverges, the given series diverges absolutely.
A: 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.
