I'm trying to show that then $\Sigma a_n$ converges iff $\Sigma 2^n a_{2^n}$ converges if $a_n$ is any monotone sequence.
I am assuming I have to prove the cases:
- $a_n$ is positive and increasing
- $a_n$ is negative and increasing
- $a_n$ is positive and decreasing
- $a_n$ is negative and decreasing
Here is what I have worked out from some of the comments below:
Assume $a_n$ is an increasing sequence. If $a_n$ is positive, then $\lim a_n \neq 0$ and we have divergence.
I don't really understand why this is. Is it because $a_n$ is unbounded and so it diverges to $\infty$? I don't know a rigorous way to prove this. I'm assuming I have to make use of the theorem that monotonic $s_n$ converges iff it is bounded.
If $a_n$ is negative, then we have that $\lim a_n = 0$
Again, I don't know how to rigorously prove this. How do I find a bound for $a_n$?
Let $s_n = a_1 + a_2 + … + a_n$ and $t_n = a_1 + 2 a_2 + … + 2^k a_{2^k}$.
For $n<2^k$, we have $s_n \geq a_1 + (a_2 + a_3) + … _ (a_{2^k}+…+a_{2^{k+1}-1}
\geq a_1 + a_2 + … + 2^{k}a_{2k} = t_k
$And so $s_n \geq t_k$.
Similar proof for $2s_{n} \leq t_k$
Someone below has already proved nonnegative decreasing case. I'm assuming that:
If $a_n$ was decreasing and negative, then $\lim a_n \neq 0$ and we have divergence.
Again, I'm not sure how to prove this.
Best Answer
Consider the series $\sum a_n $ where $(a_n)$ is decreasing and positive. Also, consider the series $\sum 2^n a_{2^n}$. Consider their partial sums $(s_n), (t_n)$
Note if $n < 2^n$, then $s_n = a_1 + a_2 + ... + a_n \leq a_1 + (a_2+a_3) + (a_4 + a_5 + a_6 + a_7) + ... + (a_{2^n} + ... + a_{2^{n+1}-1}) \leq a_1 + 2a_2 + ... + 2^n a_{2n} = t_n \implies \boxed{ s_n \leq t_n }$
If $2^n < n$, then $s_n \geq a_1 + a_2 + (a_3 + a_4) + ... + (a_{2^{n-1}+1} + ... + a_{2^n} ) \geq \frac{1}{2} a_1 + a_2 + 2 a_4 + ... + 2^{n-1}a_n = \frac{1}{2} t_n \implies \boxed{ 2s_n \geq t_n }$
Since series of nonnegative terms converges iff its sequence of partial sums are bounded, then we are done.