# If $|a_1|>1$, then the series $\sum\frac{a_1^n+\cdots+a_k^n}{n}$ diverges

Let $a_1,...,a_k$ be complex numbers different from $1$. I am trying to prove that if at least one of them has a module strictly greater than $1$, say $|a_1|>1$, then the series $\displaystyle\sum\frac{a_1^n+\cdots+a_k^n}{n}$ diverges.

I have tried to prove that $\dfrac{a_1^n+\cdots+a_k^n}{n}$ doesn't converge to $0$ as it should for the series to converge, and for that I've rewritten it as:

$$\frac{|a_1^k+\cdots+a_n^k|}{k}= \frac{|a_1|^k}{k}\Big|1+\Big(\frac{a_2}{a_1}\Big)^k +\cdots+\Big(\frac{a_n}{a_1}\Big)^k\Big|$$ I thus have to prove that the sum after the $1+\cdots,\,$ doesn't converge to $-1$, but I can't seem to do it.

• Looking at my deleted answer, it is enough to prove the claim in the case where $|a_1|=\ldots = |a_n|$. But that looks quite hard... – Gabriel Romon May 17 '18 at 18:09

We shall show that, if at least one of $|a_1|,\ldots,|a_k|$ is larger than one, then the sequence $\displaystyle b_n=\frac{1}{n}(a_1^n+\cdots+a_k^n)$ is not bounded, and hence $\displaystyle\sum_{n=1}^\infty\frac{1}{n}(a_1^n+\cdots+a_k^n)$ diverges.

Assume, without loss of generality that $$|a_1|=\cdots=|a_m|=r>s=|a_{m+1}|\ge |a_{m+2}|\ge\cdots\ge|a_k|,$$ where $r>1$.

We shall use the following lemma (its proof is postponed).

Lemma. If $w_1,\ldots,w_m\in \mathbb C$, with $|w_1|=\cdots=|w_m|=1$, then the sequence $z_n=w_1^n+\cdots+w_m^n$ does not tend to zero. Hence, there exists an $\eta>0$, such that $|z_n|\ge \eta$, for infinitely many $n$'s.

So, setting $a_1=rw_1,\ldots,a_m=rw_m$, then the Lemma provides that there exists an $\eta>0$ and infinitely many $n$'s such that $$|a_1^n+\cdots+a_m^n|=r^n|w_1^n+\cdots+w_m^n|\ge \eta r^n,$$ and hence $z_n$ is unbounded since, for infinitely many $n$'s $$|a_1^n+\cdots+a_k^n|\ge |a_1^n+\cdots+a_m^n|-|a_{m+1}^n+\cdots+a_k^n|\ge r^n|w_1^n+\cdots+w_m^n|-(k-m)s^n \\\ge \eta r^n-(k-m)s^n=\eta r^n \left(1-\frac{k-m}{\eta}\Big(\frac{s}{r}\Big)^n\right)>\frac{\eta}{2}r^n,$$ where the last inequality holds for sufficiently large $n$, as $s/r<1$.

Proof of the Lemma. Assume that $z_n\to 0$, as $n\to\infty$. The terms $w_1,\ldots,w_k$ do not have to be different. Say that there are only $\ell$ different complex numbers is the set $\{w_1,\ldots,w_k\}$, without loss of generality the $w_1,\ldots,w_\ell$, are different from each other and they appear $j_1,\ldots,j_\ell$ times, respectively (with $j_1+\cdots+j_\ell=k$). We have $$j_1w_1^n+\cdots+j_\ell w_\ell^n=z_n,\\ w_1 j_1w_1^n+\cdots+w_\ell j_\ell w_\ell^n=z_{n+1},\\ \cdots\\ w_1^{\ell-1} j_1w_1^n+\cdots+w_\ell^{\ell-1} j_\ell w_\ell^n=z_{n+\ell-1}.$$ The above is viewed as an $\ell\times\ell$ linear system with unknowns the $j_1 w_1^n,\ldots,j_\ell w_\ell^n$ and system-matrix $A=(w_i^{j-1})_{i,j=1,\ldots,\ell}$ is the Vandermonde matrix, which is invertible, as the $w_1,\ldots,w_\ell$ are different from each other. Hence $$(j_1w_1^n,\ldots,j_\ell w_\ell^n)^T=A^{-1}(z_n,\ldots,z_{n+\ell-1})^T.$$ So, if $z_n\to 0$, then so does the right-hand side of the above, and hence the left-hand side of the above. But $j_i|w_i|^n=j_i\ne 0$, for all $i=1,\ldots,\ell$. Contradiction. This concludes the proof of the Lemma.

• The lemma is false as stated, as the moduli could be less than $1$ in which case you get convergence to $0.$ – zhw. May 17 '18 at 20:33
• @zhw. You are right - I corrected formulation and the proof of the lemma. – Yiorgos S. Smyrlis May 17 '18 at 20:40

Suppose that the series is convergent. Then the power series $$f(z)=\sum_{k=1}^{\infty} \frac{a_1^k+\cdots + a_n^k}k z^k$$ is valid on $D=\{z: |z|<1\}$.

By differentiation, we have $$f'(z)=\sum_{k=1}^{\infty} (a_1^k + \cdots + a_n^k)z^{k-1}$$ is valid on $D$.

Moreover the series on the right side equals $$\frac{a_1}{1-a_1z} + \cdots +\frac{a_n}{1-a_n z}$$ on $|z|<\min \{1/|a_i| : 1\leq i\leq n\}<1$ since each term is geometric series.

This function has a pole at $1/a_1$ which is inside $D$. This is a contradiction. Thus, the series must be divergent.

• Isn't it rather $f'(z)=\sum_k^\infty k(a_1^k+...+a_n^k)z^{k-1}$? Also, I believe you have to prove each series converges before separating the sum (what if one pole and another `compensate'?). – John Do May 18 '18 at 13:56
• $k$ doesn't appear in $f'(z)$ since $f(z)$ had $k$ in the denominator. Also, the series $f'(z)$ as the sum of rational functions is originally valid on $|z|<R$ where $R$ is minimum of $1/|a_i|$. But, is also valid on $D$ due to analytic continuation. And, the poles does not compensate. – Sungjin Kim May 18 '18 at 14:00
• The above comment is under the assumption of convergence of the series. – Sungjin Kim May 18 '18 at 14:01
• This is a great solution! +1. However, you did not explain, in your answer, why you can pull apart the series as you did. It would be good to do that, since it's a big part of the proof. So, assuming $a_1$ has the largest modulus among the $a_m,$ each of the series $\sum a_m^kz^{k-1}$ is a geometric series converging in $\{|z|<1/|a_1|\}.$ In that smaller disc, we can pull apart the series as you did. We then see $\lim_{r\to 1^-}|f(r/|a_1|)| \infty,$ contradiction, since $f'$ was analytic in the unit disc. No need to speak of poles or analytic continuation. – zhw. May 19 '18 at 15:51
• @JohnDo Take a look at my comment. No need for analytic continuation ... – zhw. May 19 '18 at 15:54