Cauchy´s convergence test for Series Show whith the cauchy's convergence test for series, that the sequence :
$$ b_n = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}     $$
converges.  
I think it would be not so difficult to use the comparison test, but using the cauchy's test I find it more tricky. Also somehow I used a similar proof to the one used on the harmonic serie and showed that the serie diverges, but there must be an error there. I did: 
$$\left|\sum_{k=2n+1}^{4n} \frac{1}{k} \right| = \frac{1}{2n+1}+\frac{1}{2n+2}+\cdots+\frac{1}{4n}\geq 2n\frac{1}{4n}=\frac{1}{2}>\epsilon  $$
Do you find the error in that proof? And how do I show correctly that the cauchy's test actually smaller than every positive real number is?
Thank you for your help!
 A: The statement
$$
\left|\sum_{k=2n+1}^{4n} \frac{1}{k} \right| = \frac{1}{2n+1}+\frac{1}{2n+2}+\cdots+\frac{1}{4n}\geq 2n\frac{1}{4n}=\frac{1}{2}>\epsilon
$$ does not imply that $b_{2n}$ diverges (i.e. does not converge to $\textbf{any}$ value), but imply that $b_{2n}$ does not converge to $0$. This is the error in your proof. 
In fact Cauchy's test can be done in this way. Observe that
$$
0\leq b_{n+1}-b_n = \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1} = \frac{1}{2n+1}-\frac{1}{2n+2}\leq \frac{1}{4n(n+1)},
$$ and hence for $m>n\geq N$,
$$0\leq b_m-b_n= \sum_{k=n}^{m-1}(b_{k+1}-b_k)\leq \sum_{k=n}^{m-1}\frac{1}{4}(\frac{1}{k}-\frac{1}{k+1})\leq \frac{1}{4}(\frac{1}{m}-\frac{1}{n+1})\leq \frac{1}{4N}.$$
A: If you know some asymptotic approximation for $H_n$ it is no wonder that $b_n=H_{2n}-H_n$ converges to $\log 2$. Anyway, let us try to outline some interesting facts here. For instance, $b_n$ is a Riemann sum for $\int_{0}^{1}\frac{dx}{1+x}$: since $\frac{1}{1+x}$ is a convex function on $[0,1]$, Karamata's inequality ensures that $\{b_n\}_{n\geq 1}$ is increasing. This can be also checked with elementary manipulations:
$$ b_{n+1}-b_n = \frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)}>0.$$
This identity also gives
$$ \lim_{n\to +\infty}b_n = \lim_{n\to +\infty}\sum_{m=0}^n\frac{1}{(2m+1)(2m+2)}=\sum_{m\geq 0}\frac{1}{(2n+1)(2m+2)}. $$
$b_n$ has a simple integral representation: since $\frac{1}{b+1}=\int_{0}^{1}x^{b}\,dx$,
$$\begin{eqnarray*} b_n = H_{2n}-H_n &=& \sum_{m=0}^{n}\int_{0}^{1}\left(x^{2m}-x^{2m+1}\right)\,dx=\int_{0}^{1}\frac{1-x^{2m+2}}{1+x}\,dx\\&=&\log(2)-\int_{0}^{1}x^{2n}\frac{x^2 dx}{1+x}.\end{eqnarray*}$$
The error term $\log(2)-b_n=\int_{0}^{1}x^{2n}\frac{x^2\,dx}{1+x}$ is trivially non-negative and decreasing to zero (by the dominated convergence theorem), bounded between $\frac{1}{4n+6}$ and $\frac{1}{2n+3}$, log-convex (and so convex) since it is a moment: the function $f(s)=\int_{0}^{1} x^s \frac{x^2\,dx}{1+x}$ is clearly continuous on $[-1,+\infty)$ and the midpoint-log-convexity of $f(s)$ is ensured by the Cauchy-Schwarz inequality:
$$\left[\int_{0}^{1}x^{\frac{s+t}{2}}\frac{x^2\,dx}{1+x}\right]^2 \leq \int_{0}^{1}x^s\frac{x^2\,dx}{1+x}\int_{0}^{1}x^t\frac{x^2\,dx}{1+x} $$
is equivalent to $\log f\left(\frac{s+t}{2}\right)\leq \frac{\log f(s)+\log f(t)}{2}.$
