TITLE: If $|a_1|>1$, then the series $\sum\frac{a_1^n+\cdots+a_k^n}{n}$ diverges
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: 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.