Convergence of $a_n=1+1/5+1/9+\ldots+\frac{1}{4n-3}$ Show that the sequence $$a_n=1+1/5+1/9+\ldots+\frac{1}{4n-3}$$ does not converge but the sequence $b_n=\frac{a_n}{n}$ converges.
I can show the first part. For the second part, will it be sufficient to just show that $$\lim \frac{a_n}{n}=0?$$
 A: Just note that

$$ a_n = \sum_{i=1}^{n} \frac{1}{4i-3} \sim \int_{1}^{n+1} \frac{dx}{4x-3}=\dots.  $$

For the other one 

$$\lim_{n\to \infty} \frac{a_n}{n} \sim \lim_{n\to \infty}\frac{\int_{1}^{n+1} \frac{dx}{4x-3}}{n},  $$

you can use L'hopital's rule.
A: Apply Cesaro-Stolz's theorem for $x_n = \dfrac{1}{4n-3}$, and note that : $b_n = \dfrac{x_1 + x_2 +...+ x_n}{n}$
A: An elementary proof: Let $b_n = a_n/n$. Then 
$$
b_{n+1} = \frac{n}{n+1}b_n + \frac{1}{4n+1}.
$$
First prove by induction that $\frac{n+1}{n(4n+1)}< b_n$ and then deduce that $0<b_{n+1}< b_n$. Therefore the sequence is bounded below and decreasing, so it is convergent.
A: note that $a_{{n}}=\sum _{k=1}^{n}{ \frac{1}{4k-3}}$, then
$\sum _{k=1}^{n}{ \frac{1}{4k-3}}=\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{k-\frac{3}{4}}}$
$>\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{k}}$
and the last series is the harmonic series, which does not converge. Therefore $a_{{n}}$ does not converge. Similarily, $b_{{n}}=\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{n(k-\frac{3}{4})}}$, and
$\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{n(k-\frac{3}{4})}}<\frac{1}{4}\sum _{k=1}^{n}{ \frac{1}{(k-\frac{3}{4})^2}}=\frac{1}{4}\sum _{k=2}^{n}{ \frac{1}{(k-\frac{3}{4})^2}}+4<\frac{1}{4}\sum_{k=1}^{n}{\frac{1}{k^2}}+4$
and the last series converges. Therefore $b_{{n}}$ also converges.
