Dominated convergence of series of powers

Question

Let $(a_i) \in \mathbb{C}$ be a sequence of complex numbers s.t. $|a_i| < 1$ for every $i$ and

$$\sum_{i=1}^{\infty} a_i^k \text{ is convergent for all } k \in \mathbb{N} $$

Then, must it be the case that $\lim_k \sum_{i=1}^{\infty}a_i^k = 0 $?

This is a question that arose for me while tackling a competition-style problem, for which a proof of this would give me a full solution. However, I think it is of interest within its own right as well.

Note that if the series corresponding to $(a_i)$ were absolutely convergent, then this would follow essentially from the dominated convergence theorem. It may also be observed that, in general, there is no need for there to exist $k$ s.t. $\sum a_i^k$ is absolutely convergent, since $\bigcup_p \ell^p \neq c_0$, where $c_0$ is the space of all sequences converging to zero.

Answer 1

We can construct the counterexample as follows. Let the sequence $(a_i)_{i\in\mathbb N}$ consists of consecutive blocks. For any natural $m$ let $p_m$ be the $m$-th prime number, $\xi_m=\exp{\frac{2\pi i}{p_m}}$, $n_m=p_m^{2p_m}$, and the $m$-th block be $\frac 1{p_m^2}\left(\xi_m^0,\xi_m^1,\dots,\xi_m^{n_m-1}\right)$. Then for each prime $p$, $$\sum_{i=1}^\infty a_i^p=1.$$ In fact, for each natural $k\ge 2$, $$\sum_{i=1}^\infty a_i^k=\sum \{p^{2(p-k)}: p \mbox{ is a prime divisor of }k\}$$ and $$\sum_{i=1}^\infty a_i=0.$$

Answer 2

Not an answer, and I like the existing answer, but I thought I could try and give some intuition. I hope it’s useful and not too off topic.

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One of the most general theorems for exchanging limits and series involves the so-called “tightness” condition. That is, given coefficients $a_i^{(k)}$ (where the superscript is just an index, not a power), it holds $$ \lim_{k\to\infty}\sum_{i=1}^\infty a_i^{(k)}= \sum_{i=1}^\infty\lim_{k\to\infty}a_i^{(k)} $$ if the following condition holds: for every $\varepsilon>0$ there exists $M_\varepsilon$ such that $$ \sup_{k} \sum_{i=M_\varepsilon}^\infty|a_i^{(k)}|<\varepsilon. $$ This condition is called tightness of the sequence of series, and essentially says that “the mass of the sequence of coefficients does not escape at infinity”: in that case, switching between limit and series is allowed. This is a generalization of dominated convergence theorem (a general version for integrals exists as well, it’s called Vitali convergence theorem).

A classical example of when exchanging limit and series fails is of course that of $a^{(k)}_i=\delta_{i,k}$: the series $\sum_i a^{(k)}_i$ equals $1$ for all $k$, but the coefficients converge pointwise to zero as $k\to\infty$. This is a nice case to visualize what one means by “mass escaping at infinity”: the support of the coefficients literally escapes to infinity. This is sometimes called the “escaping square” in a similar context with integrals instead of series.

You can build a similar counterexample also with coefficients going to zero very fast, and uniformly in the index $i$: $a_i^{(k)}=2^{-k}$ for $i=2^k+1,\dots,2^{k+1}$ and zero otherwise. I usually call this “escaping peanut butter⁽¹⁾”.

Of course the above theorem applies only to absolutely convergent sequences, so it does not apply to your problem. I mentioned that to add some intuition on the fact that whenever the mass of the series is allowed to escape at infinity, then you might expect to see the limit of the series to be different from the series of the pointwise limits of the coefficients, unless you have some other conditions like monotonicity (e.g. for real sequences of series whose terms decrease monotonically). In your case, if $\sum a^k_i$ does not converge absolutely, you have literally infinite mass to work with without conditions at infinity that are uniform in the index $(k)$, and you can expect to be able to prepare the coefficients in the right way so that you see something moving to infinity like an escaping peanut butter.

In your specific problem, it is of course far from trivial to tell whether that can actually be done and how to do that in practice, and the answer of @Alex_Ravsky gives a very nice example on how to do that.

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^{(1) you know, when you are spreading peanut butter on bread and you say “ok, I have eaten too much”, but you don’t want to stop eating, so you take another slice of bread and spread a thinner film of p.b. to convince yourself that you are eating less, but you also take twice the amount of bread you took before, so you are actually eating the same amount of p.b., and this way your belly escapes to infinity.}