Proving that if the partial sums are bounded (i.e. $|\sum_n^N a_n|$ is bounded) and that $\sum |a_n|^2$ converges, then $\sum a_n$ converges. I am wondering if the following is true and if so need help proving it. If the series of partial sums is bounded, that is $|\sum_{n=1}^N a_n|$ is a bounded sequence indexed by $N$ and that $\sum_{n=1}^{\infty} |a_n|^2$ converges, then $\sum_{n=1}^{\infty} a_n$ converges. The $a_n$ are complex.
Thanks!
 A: Look at 
$$1-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{8}-\frac{1}{8}-\frac{1}{8}-\frac{1}{8}-\frac{1}{8}-\frac{1}{8}-\frac{1}{8}-\frac{1}{8}+\frac{1}{16}+\cdots.$$
The series does not converge. But $1$ is an upper bound for the partial sums, and $0$ is a lower bound, and $\sum a_n^2=2$.
A: This is not true. Maybe $$\sum_{n=1}^Na_n=\mathrm{e}^{if(N)}$$ so that $\left|\sum_{n=1}^Na_n\right|$ is bounded (equals $1$). We can easily find $f(N)$ so that $\sum_{n=1}^Na_n$ diverges, but $\sum_{n=1}^N|a_n|^2$ converges. Basically, we want the partial sums to walk along the unit circle taking steps whose absolute value is roughly $1/n$. We could have $f(N)=\ln(N)$ for instance.
To clarify, for $n>1$ this example has 
$$\begin{align}
a_n&=\sum_{j=1}^na_j-\sum_{j=1}^{n-1}a_j\\
&=\mathrm{e}^{if(n)}-\mathrm{e}^{if(n-1)}\\
&=\mathrm{e}^{i\ln(n)}-\mathrm{e}^{i\ln(n-1)}\\
&=\mathrm{e}^{i\ln(n)}\left(1-\mathrm{e}^{\ln(n-1)/\ln(n)}\right)\\
&=\mathrm{e}^{i\ln(n)}\left(1-(n-1)^{1/\ln(n)}\right)
\end{align}$$
This difference has absolute value $1-(n-1)^{1/\ln(n)}$. This is smaller than $\frac{1}{n}$ (verified below), so $|a_n|^2&lt\frac{1}{n^2}$, implying $\sum|a_n|^2$ converges.
To see the inequality holds:
$$\begin{align}
&&(n-1)^{\ln(n)}&&lt n^{\ln(n)}\\
&\implies&(n-1)^{\ln(n)}&&lt(n-1)\cdot n^{\ln(n)}\\
&\implies&\left(1-\frac{1}{n}\right)^{\ln(n)} & &lt(n-1)\\
&\implies&1-\frac{1}{n}&&lt(n-1)^{1/\ln(n)}\\
&\implies&1-(n-1)^{1/\ln(n)} & &lt\frac{1}{n}
\end{align}$$
