Show that $\sum\limits_{k=1}^\infty\left| \left(1-\frac{1}{2k}\right)\left(\frac{k+1}{k}\right)^{1/2}-1\right|$ converges. Show that $\sum\limits_{k=1}^\infty\left| \left(1-\frac{1}{2k}\right)\left(\frac{k+1}{k}\right)^{1/2}-1\right|$ converges.
I was thinking that if I show that $\left| \left(1-\frac{1}{2k}\right)\left(\frac{k+1}{k}\right)^{1/2}-1\right|$ is less than or equal to the kth term of some geometric series then I would be done, but I am having trouble thinking of such a series so I am questioning if this is even the right approach?
 A: Note that from the generalized binomial theorem, we can write
$$\begin{align}\left(1 -\frac1{2k}\right)\left(1 +\frac1{k}\right)^{1/2}-1&=\left(1 -\frac1{2k}\right)\left(1 +\frac1{2k}+O\left(\frac1{k^2}\right)\right)-1\\\\
&=O\left(\frac{1}{k^2}\right)
\end{align}$$
Then, compare the sum of this to the series $\sum_{k=1}^\infty\frac1{k^2}=\frac{\pi^2}{6}$
A: Using $(a-b)(a+b) = a^2-b^2$, show first that each term is less than or equal to $| ( 1 - \frac 1 {2k} )^2 (\frac{k+1}k) -1 |$. Then do the algebra on this term - you will see it simplifies to $\frac{3k-1}{4k^3}$. This is less than $\frac 1 {k^2}$. Use the fact that $\sum \frac 1 {k^2}$ converges.
A: May be, you could consider $$u_k=\left(1-\frac{1}{2 k}\right) \sqrt{\frac{k+1}{k}}-1$$ For the time being, set $k=\frac 1x$ and what you look at is $$(1-\frac x2)\sqrt{1+x}-1$$ Using Taylor expansion around $x=0$, you have $$(1-\frac x2)\sqrt{1+x}-1=-\frac{3 x^2}{8}+\frac{x^3}{8}+O\left(x^4\right)$$ Back to $k$ $$u_k=-\frac{3}{8 k^2}+\frac{1}{8 k^3}+O\left(\frac{1}{k^4}\right)$$ I am sure that you can take it from here.
