Show that $\sum_{n=0}^{\infty} a_{f(n)}$ is also an absolutely convergent series.

Question

Let $\sum_{n=0}^{\infty} a_n$ be an absolutely convergent series of real numbers. Let $f:N \to N$ be an increasing function (i.e., $f(n+1) > f(n)$ for all $n\in N$). Show that $\sum_{n=0}^{\infty} a_{f(n)}$ is also an absolutely convergent series.

Hint: Try to compare each partial sum of $\sum_{n=0}^{\infty} a_{f(n)}$ with a (slightly different) partial sum of $\sum_{n=0}^{\infty} a_n.$

My attempt:

If $\sum_{n=0}^{\infty} a_n$ is absolutely convergent, then by definition, $\sum_{n=0}^{\infty} a_n$ and $\sum_{n=0}^{\infty} |a_n|$ converges. By the Proposition of "Rearrangement of series," Let: $$L= \sum_{n=0}^{\infty} |a_n|.$$ Then, $\sum_{n=0}^{\infty} a_{f(n)}$ converges to $L$. Let: $$L'=\sum_{n=0}^{\infty} a_n.$$ Then, we want to show that $\sum_{n=0}^{\infty} a_{f(n)}$ also converges to $L'$. In other words, given any $\epsilon>0$, we have to find an $M$ such that $\sum_{n=0}^{M'} a_{f(n)}$ is $\epsilon$-close to $L'$ for every $M'>M$.

Answer 1

I think the proof is easier. Let me sketch my thoughts. If $f$ is strictly increasing, then the image of $f$ is some subset of $\mathbb N$ which may be all of $\mathbb N$ or could be smaller. (To see this, observe that $f(1) = n$ need not imply $f(2) = n+1$ - it could be larger, so there could be a gap which we never come back to since this function is strictly increasing). If $f$ is the identity, then there's nothing to prove. Otherwise, $f$ amounts to a deletion of some collection of terms, which, upon taking absolute values, is strictly smaller than the absolute value of the series you started with.

Answer 2

Observe that $$\sum_{k=0}^n |a_{f(k)}| \le \sum_{k=0}^{f(n)} |a_k| \le L$$ so that the absolute series has bounded partial sums and is thus absolutely convergent.

Answer 3

$f$ is injective. Any injective function works. In fact, consider

$$\sum_{n=1}^N a_{f(n)}$$

and let $M$ be the biggest value taken on by $f(n)$ i.e. $M = \max_{ n \le N} f(n)$

Then $$\sum_{n=1}^N a_{f(n)} \le \sum_{n=1}^M |a_{n}| \le L$$

where $L = \sum_{n=1}^\infty |a_n|$. Then taking the limit as $N \to \infty$ you get the result. Notice that we only used (in the first inequality) the property that $f$ is injective.