Prove that sequence $a_n/n$ is convergent for $a_n = 1+\frac15+\frac19... + \frac1{4n-3}$ Show that the sequence $\langle a_n\rangle$ where 

$a_n = 1+\frac{1}{5}+\frac{1}{9}... + \frac{1}{4n-3}$

does not converge, whereas $\langle b_n\rangle$ defined by $b_n = \frac{1}{n}  a_n$ converges to 0.

My Approach: I was able to prove that $\langle a_n\rangle$ does not converge by showing that it is not a Cauchy sequence but not able to prove that $\langle b_n\rangle$ is converging sequence. Is there a theorem related to this?.
I completely misread the question at first. Let's try again. We know that for $k\geq1$, we have $4k-3\geq k$, and hence 
$$
\sum_{k=1}^n\frac{1}{4k-3}\leq \sum_{k=1}^n\frac{1}{k}.
$$
The sum on the right gives the harmonic numbers, $H_n$, which are approximated asymptotically by $\ln(n)$ (given by a Riemann approximation of $\int_1^n\frac{1}{x}dx$). Thus since $$\lim_{n\to\infty}\frac{\ln(n)}{n}=0,$$
we may apply the squeeze theorem to 
$$0\leq \frac{1}{n}\sum_{k=1}^n\frac{1}{4k-3}\leq \frac{C+\ln(n)}{n}$$
(where $C$ is an arbitrarily large, fixed constant to make the inequality hold). Thus we have $b_n\to0$.
$$1+1/5+1/9+...\geq 1/2+1/6+1/10+...=$$ $$=(1/2)(1+1/3+1/5+...)\geq (1/2)(1/2+1/4+1/6+...)=$$ $$=(1/4)(1+1/2+1/3+...).$$ Let $s_n=\sum_{j=1}^n1/j.$ We have  $$s_2=3/2.$$ $$s_4=s_2+(1/3+1/4)&gt;s_2+2(1/4)=2.$$ $$s_8=s_4+(1/5+...+1/8)&gt;s_4+4(1/8)&gt;2+1/2=5/2.$$ $$s_{16}=s_8+(1/9+...+1/16)&gt;s_8+8(1/16)&gt;5/2+1/2=3.$$ Et cetera.