3
$\begingroup$

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!

$\endgroup$
6
$\begingroup$

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 $\sum a_n^2=2$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.