# Divergence of subsequences

How can we show that if $f:\mathbb{N} \rightarrow \mathbb{N}$ with $f(n+1) \geq f(n)>1$ for all $n\ge 1$ and $\sum_{n = 1}^\infty \frac{1}{f(n)} = \infty$, then for all integers $k>1$ we have $\sum_{n = 1}^\infty \frac{1}{f(kn)} = \infty$?

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If the sum is finite, then so is $\sum \frac{1}{f(kn+r)}$ where $0\le r \le k-1$. – André Nicolas Nov 14 '11 at 7:43

$$\frac{k}{f(kn)}\geqslant\frac1{f(kn+1)}+\frac1{f(kn+2)}+\cdots+\frac1{f(kn+k)}.$$