# $\sum a_n$ converges absolutely, does $\sum (a_n + \cdots + a_n^n)$ converge

Suppose $\sum_{n=1}^\infty a_n$ converges absolutely. Does this imply that the series $$\sum_{n=1}^\infty (a_n + \cdots + a_n^n)$$ converges?

I believe the answer is yes, but I can't figure out how to prove it. Any help would be appreciated.

Thanks.

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Could you clarify? Is this a sum $\sum^{\infty}_{n=1} a_n^n$ or something else? – Karolis Juodelė Jul 26 '12 at 6:29
No, its $\sum_{n=1}^\infty \sum_{m=1}^n a_n^m$. – Galois Jul 26 '12 at 6:31
@Karolis: No, it's $(a_1) + (a_2 + a_2^2) + (a_3 + a_3^2 + a_3^3) + \ldots$. – mixedmath Jul 26 '12 at 6:31
@mixedmath precisely – Galois Jul 26 '12 at 6:32
Very nice problem! (+1) – user 1618033 Jul 26 '12 at 10:49

## 2 Answers

After a while, $|a_k| \lt \frac{1}{2}$.

From that point on, $$|a_k+a_k^2+a_k^3+\cdots +a_k^k| \le |a_k|+\frac{1}{2}|a_k|+\frac{1}{4}|a_k|+\cdots+\frac{1}{2^{k-1}}|a_k|\le 2|a_k|.$$ So by comparison our series converges absolutely.

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Nice thanks. ${}$ – Galois Jul 26 '12 at 6:40
@André Nicolas: i wanted to post the same answer when i saw it. Now i think of a different approach. (+1) – user 1618033 Jul 26 '12 at 10:50

Another way to see this is to notice that each term is less than $a_n + a_n^2 + a_n^3 + ...$ which, when $a_n < 1$ is $\frac{a_n}{1-a_n}$. When $a_n < \frac{1}{2}$ then $\frac{a_n}{1-a_n} < 2a_n$.

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