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Related to my previous question, consider $$f(s)=\sum_{k=1}^{\infty} \exp(-s(-2)^k)$$where $s\in\mathbb{C}$ is a complex number. According to the Willie Wong's comment, $f(s)$ diverges when $\Re\{s\} \not = 0$ and my numerical approximation using Mathematica supports this claim (honestly, I couldn't prove it for myself). My main interest is the case $\Re\{s\} = 0$ and the partial sums suggests that $f(s)$ converges when $s$ satisfies the conditions $\Im\{s\}\lt 1, \Re\{s\} = 0, s \not = 0$. In the other case, i.e. $\Im\{s\}\ge 1, \Re\{s\} = 0$, Mathematica returns "Overflow occurred in computation". So the main question is when $f(s)$ converges?

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    $\begingroup$ The condition $\Im\{s\}\lt 1, \Re\{s\} = 0, s\neq0$ cannot guarantee $f(s)$ to converge. Consider $s=i\pi/4$, then $$\sum_{k=3}^\infty e^{-s(-2)^k}=\sum_{k=3}^\infty e^{-i\pi(-2)^{k-2}}=\sum_{k=1}^\infty(e^{-i\pi})^{(-2)^k}=\sum_{k=1}^\infty(-1)^{(-2)^k}=\sum_{k=1}^\infty1=\infty$$ $\endgroup$
    – Mengchun Zhang
    Commented Feb 10 at 22:05
  • $\begingroup$ @MengchunZhang You are right, thanks. Do you have any idea for the main question? I mean for what values of $s$ the summation converges? $\endgroup$
    – S.H.W
    Commented Feb 10 at 22:24
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    $\begingroup$ Thanks for the reply, in fact, I think $f(s)$ diverges for all $s\in\mathbb C$. If $\Re\{s\}\neq0$, then Willie Wong’s comment showed it diverges; if $\Re\{s\}=0$, then the summand has modulus $1$, i.e. $|\exp(-s(-2)^k))|=1$ for all $k$, so the summand doesn’t tend to $0$, meaning the series doesn’t converge. $\endgroup$
    – Mengchun Zhang
    Commented Feb 11 at 0:53
  • $\begingroup$ @MengchunZhang Thanks. Would you turn your comment into an answer? Actually I couldn't prove to myself that when $\Re\{s\} \not = 0$, $$|\sum_{m=1}^{k-1} \exp(-s(-2)^m)| \le |\exp(-s (-2)^k) + \exp(-s (-2)^{k+1})|$$ holds but it seems to me that also in this case modulus of summand doesn't tend to $0$ since we have $$|\exp(-(a+bi)(-2)^k)| = \exp(-a(-2)^k)$$ and it's possible to find a subsequence which diverges to infinity. Is this reasoning correct? $\endgroup$
    – S.H.W
    Commented Feb 11 at 11:51

1 Answer 1

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Among American college students, the following is sometimes called the "n-th term test":

A necessary condition for a series $\sum a_n$ to converge is that $\lim a_n = 0$.

A fortiori, if $\sum a_n$ converges then $\lim |a_n| = 0$ (here we allow $a_n$ to take complex values). Apply this to our series:

Given $s\in \mathbb{C}$:

  • If $\Re(s) > 0$, then as $k$ increases among the odd numbers, $|\exp(-s(-2)^k)|$ diverges to $+\infty$.
  • If $\Re(s) < 0$, then as $k$ increases among the even numbers, $|\exp(-s(-2)^k)|$ diverges to $-\infty$.
  • If $\Re(s) = 0$, then $|\exp(-s(-2)^k)| = 1$ as the argument to $\exp$ is purely imaginary.

Hence $\sum \exp(-s(-2)^k)$ cannot converge for any $s\in \mathbb{C}$.

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