Prove that $\sum a_n$ converges iff $\sum 2^n a_{2^n}$ converges, but with another function instead of $2^{n}$. Let $\{a_n\}$ be a positive, decreasing sequence of real numbers.
I know how to prove that $\sum_{n=1}^\infty a_n$ converges iff $\sum_{n=1}^\infty 2^n a_{2^n}$ converges, but is it possible that this is true for another function from naturals to naturals instead of $2^n$? In case there is no function why is it?
 A: Yes, that test generalizes.  First of all, observe that passage to the subsequence $a_{2^n}$ together with a weighting by $2^n$ is a discrete analogoue of integration by substitution on $\int f(x)\,dx$ using the substitution $x = e^u$, from which we get $\int f(x)\,dx = \int f(e^u)e^u\,du$.
The fact that $e^u$ is its own derivative (w.r.t. $u$) is analogous to the sequence $2^n$ being its own discrete difference ($2^{n+1} - 2^n = 2^n$), which suggests replacing the weighting factor $2^n$ in the convergence test with a discrete difference, and there is the following result: under suitable conditions on a sequence $u(n)$, $\sum_{n \geq 1} a_n$ with nonnegative non-increasing terms $a_n$ converges if and only if  $\sum_{n \geq 1} a_{u(n)}(\Delta u)(n)$ converges, where $(\Delta u)(n) = u(n+1) - u(n))$ for all $n \geq 1$. There is no claim that the two series are equal when they both converge, just like in the special case $u(n) = 2^n$ that you describe, so it is not quite like integration by substitution. But the analogy is still striking.
The test you write about is called Cauchy's condensation test, and the bottom of the Wikipedia page on it here states the generalization I describe above and includes some precise conditions on $u(n)$ under which you can carry out this "summation by substitution".
