Ytukyg

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:
$$left|sum_{k=2n+1}^{4n} frac{1}{k} right| = frac{1}{2n+1}+frac{1}{2n+2}+cdots+frac{1}{4n}geq 2nfrac{1}{4n}=frac{1}{2}>epsilon $$
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!

The statement
$$
left|sum_{k=2n+1}^{4n} frac{1}{k} right| = frac{1}{2n+1}+frac{1}{2n+2}+cdots+frac{1}{4n}geq 2nfrac{1}{4n}=frac{1}{2}>epsilon
$$ 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
$$
0leq b_{n+1}-b_n = frac{1}{2n+1}+frac{1}{2n+2}-frac{1}{n+1} = frac{1}{2n+1}-frac{1}{2n+2}leq frac{1}{4n(n+1)},
$$ and hence for $m>ngeq N$,
$$0leq b_m-b_n= sum_{k=n}^{m-1}(b_{k+1}-b_k)leq sum_{k=n}^{m-1}frac{1}{4}(frac{1}{k}-frac{1}{k+1})leq frac{1}{4}(frac{1}{m}-frac{1}{n+1})leq frac{1}{4N}.$$

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 ${b_n}_{ngeq 1}$ 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
$$ lim_{nto +infty}b_n = lim_{nto +infty}sum_{m=0}^nfrac{1}{(2m+1)(2m+2)}=sum_{mgeq 0}frac{1}{(2n+1)(2m+2)}. $$
$b_n$ 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:

$$left[int_{0}^{1}x^{frac{s+t}{2}}frac{x^2,dx}{1+x}right]^2 leq int_{0}^{1}x^sfrac{x^2,dx}{1+x}int_{0}^{1}x^tfrac{x^2,dx}{1+x} $$
is equivalent to $log fleft(frac{s+t}{2}right)leq frac{log f(s)+log f(t)}{2}.$