How to prove that the following sequence is a Cauchy sequence? Show that the sequence $a_n$ defined as $a_n=1+1/5+1/9+...+1/(4n-3)$ does not converge whereas $b_n$=$\frac{a_n}{n}$ converges to $0$.
While I was able to do the first part by proving that $a_n$ is a not Cauchy sequence I need help proving the second part.
My attempt: Let $p=2m$ and $q=m$, so that $p,q\geq m$.
Consider $|f_p-f_q|=|f_{2m}-f_m|=|\frac{1}{n}[\frac{1}{4m+1}+\frac{1}{4m+5}...+\frac{1}{4m+4m-3}]|\leq |\frac{1}{n}[\frac{1}{4m}+\frac{1}{4m}+...+\frac{1}{4m}]|= \frac{1}{4n}$
So, $|f_p-f_q|\leq \frac{1}{4mn}$
How should I proceed further?
Thanks in advance!!
 A: For any $n \geq 2$ we have that
$$\frac{1}{4n-3} \leq \int_{n-1}^n \frac{1}{4x-3} \: dx.$$
Which means that
$$\sum_{k=2}^n \frac{1}{4k-3} \leq \int_1^n \frac{1}{4x-3} \: dx= \frac{1}{4}\log(4n-3)$$
Using this we get that
$$\frac{a_n}{n} \leq \frac{1+\frac{1}{4}\log(4n-3)}{n}$$
Using for instance L'Hôpitals rule, you may prove that
$$\lim_{n\rightarrow \infty} \frac{1+\frac{1}{4}\log(4n-3)}{n} = 0,$$
and conclude that $\lim_{n\rightarrow \infty} b_n = 0$. Note also that we could use the calculations above to conclude that
$$|b_n-b_m| \leq |b_n| + |b_m| \leq \frac{1+\frac{1}{4}\log(4n-3)}{n} + \frac{1+\frac{1}{4}\log(4m-3)}{m},$$
where the righthand side goes to $0$ as $n,m\rightarrow \infty$, proving that the sequence is cauchy.
A: Hint
$$
\frac{1}{4k+1}< \frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k+1}-\sqrt{k} \\
a_n= \sum_{k=0}^{n-1}\frac{1}{4k+1}\leq \sum_{k=0}^{n-1}( \sqrt{k+1}-\sqrt{k} )
$$
A: Make things simple using asymptotic equivalence:

*

*$\dfrac1{4n-3}\sim_ \infty\dfrac1{4n}=\frac 14 H_n$, where $H_n$ is the general term of the (divergent) harmonic series. Therefore $\sum_n a_n$ diverges.

*As a consequence, $\dfrac{a_n}n\sim_\infty\dfrac14\dfrac 1{n^2}$, a (convergent) $p$-series, so that $\sum\limits_n\dfrac{a_n}n$ is also convergent.

