Cauchy condensation test The following is Cauchy condensation test:
If $b_n \ge 0$ and $b_n \ge b_{n+1}$ then $\sum b_n$ converges if and only if $\sum 2^n b_{2^n}$ converges.
Is the following modified Cauchy condensation test possible:
Let $N \in \mathbb N$. Then if $b_n \ge 0$ and $b_n \ge b_{n+1}$ then $\sum b_n$ converges if and only if $\sum N^n b_{N^n}$ converges?

There are $N^{n+1}-N^n=N^n(N-1)$ sequence elements in the segment starting at index $N^n$ and ending at index $N^{n+1}-1$, and $$b_{N^{n+1}}\le b_k\le b_{N^n}$$ for all indices $k$ in that segment, so that 
$$(1-\tfrac1N)\sum_n N^{n+1} b_{N^{n+1}}\le\sum_k b_k\le (N-1)\sum_n N^n b_{N^{n}}.$$
Which proves the statement of the condensation test also in this generalized case. However, I never came upon a case where $N>2$ would have been useful.
