<s>[INST] 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 $$\lim \frac{a_n}{n}=0?$$
 [/INST] 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.
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