Convergence of a series given that the series with a specific parentheses insertion converges

Question

I am given a sequence $(a_n)_n$ such that $|a_n|\le\frac{1}{n}$ and the sequence $b_k=\sum_{n=k^2}^{(k+1)^2-1}a_n$ and I am given that the series $\sum_k b_k$ converges. I need to show that the original series $\sum_n a_n$ also converges. I've tried to find a nice formula for the difference of the partial sums of $\sum b_k$ and $\sum a_n$ but couldn't find something computable. Would appreciate any help.

Answer 1

Note that for $k^2\le m\le k^2+2k$, $$ \begin{align} \left|\,\sum_{j=k^2}^{m-1}a_j\,\right|+\left|\,\sum_{j=m}^{k^2+2k}a_j\,\right| &\le\sum_{j=k^2}^{k^2+2k}|a_j|\tag{1a}\\ &\le\sum_{j=k^2}^{k^2+2k}\frac1j\tag{1b}\\[3pt] &\le\frac{2k+1}{k^2}\tag{1c} \end{align} $$ Thus, pick an $\epsilon\gt0$, choose an $N$ so that for $m\gt n\ge N$, $$ \left|\,\sum_{k=n+1}^{m-1}b_k\,\right|\le\epsilon/3\tag2 $$ and $$ \frac{2n+1}{n^2}\le\epsilon/3\tag3 $$

Then for any $j\gt i\ge N^2$, there are $n\le m$ so that $n^2\le i\lt(n+1)^2$ and $m^2\le j\lt(m+1)^2$

If $n\lt m$, then $$ \begin{align} \left|\,\sum_{k=i}^{j-1}a_k\,\right| &=\left|\,\sum_{k=i}^{n^2+2n}a_k+\sum_{k=n+1}^{m-1}b_k+\sum_{k=m^2}^{j-1}a_k\,\right|\tag{4a}\\ &\le\frac{2n+1}{n^2}+\epsilon/3+\frac{2m+1}{m^2}\tag{4b}\\[9pt] &\le\epsilon\tag{4c} \end{align} $$ If $n=m$, then the second and third sums above are $0$.

Thus, the series for $a_k$ is Cauchy.