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.



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 $\sum a_n^2=2$.

  • $\begingroup$ Thanks. Just a follow up, what if I also know the partial sums of $|\sum|a_n||$ is bounded. Does that imply $\sum a_n$ converges, or does that imply the even stronger statement that $\sum|a_n|$ converges or neither? $\endgroup$ – Steven-Owen Apr 15 '12 at 19:59
  • $\begingroup$ If partial sums $\sum_{k=1}^n |a_k|$ are bounded, then $\sum{k=1}^\infty |a_k|$ converges (non-decreasing sequence which is bounded above has a limit), and so $\sum a_k$ converges. Don't need anything about $\sum a_k^2$. $\endgroup$ – André Nicolas Apr 15 '12 at 20:09
  • $\begingroup$ So the series is absolutely convergent? $\endgroup$ – Steven-Owen Apr 15 '12 at 20:11
  • $\begingroup$ The condition you mention in your first comment is exactly absolute convergence. $\endgroup$ – André Nicolas Apr 15 '12 at 20:20

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 $|a_n|^2<\frac{1}{n^2}$, implying $\sum|a_n|^2$ converges.

To see the inequality holds: $$\begin{align} &&(n-1)^{\ln(n)}&< n^{\ln(n)}\\ &\implies&(n-1)^{\ln(n)}&<(n-1)\cdot n^{\ln(n)}\\ &\implies&\left(1-\frac{1}{n}\right)^{\ln(n)} & <(n-1)\\ &\implies&1-\frac{1}{n}&<(n-1)^{1/\ln(n)}\\ &\implies&1-(n-1)^{1/\ln(n)} & <\frac{1}{n} \end{align}$$


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