Comparison $\sum_{k=0}^{\infty} a_k$, $\sum_{k=0}^{\infty} a_{n_k}$ with $\lim_{k\rightarrow\infty}\frac{n_k}{k}=l<\infty$

We all know the Limit comparison test:

Suppose that we have two series $\sum_{k=0}^{\infty} a_k$, $\sum_{k=0}^{\infty} b_{k}$ with $a_k, b_k \geq 0$ for all $k$. Then if $\lim_{k \to \infty} \frac{a_k}{b_k} = c$ with $0 < c < \infty$ then either both series converge or both series diverge.

Now, we suppose that there exists a sequence $n_k$ such that $\lim_{k\rightarrow\infty}\frac{n_k}{k}=l<\infty$. Then, can we conclude that either both series converge or both series diverge? Is it an equivalent form of the Limit comparison test?

• Your question is not clear. Suppose there is a sequence $n_k$ with some property. What is the relation of $n_k$ with $a_k$ and/or $b_k$? – Emanuele Paolini Dec 16 '13 at 8:31

If $\sum a_k$ (with all $a_k\ge 0$) converges then so does $\sum a_{n_k}$, but we cannot conclude inthe other direction. For example let $n_k=2k$, $a_k=1-(-1)^k$.

• Thanks. Can you get a proof please for direction from $\sum a_k$ to $\sum a_{n_k}$? – Mark Dec 7 '13 at 9:30
• @Mark there are counterexamples even for the other direction: consider $a_k=(-1)^n/n$ and $n_k=2k$. – Norbert Dec 10 '13 at 12:34
• @Hagen Von Eitzen see my comment above – Norbert Dec 10 '13 at 12:35
• @Norbert Oh, I was assuming that the condition $a_k\ge 0$ from the quoted comparison test was assumed for the conjectured test as well ... – Hagen von Eitzen Dec 10 '13 at 19:40

This looks close to the Cauchy condensation test. For it to work, you would need the sequence $(a_k)$ monotonically decresing towards zero as well.

Then

$\sum_k (n_k-n_{k-1})a_{n_k}\le \sum_k a_k\le\sum_k (n_{k+1}-n_k)a_{n_k}$

However, it is not clear to me if the size of $(n_{k+1}-n_k)$ can be meaningfully bounded by $l$.