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Assume that $n_k$ is a sequence of natural numbers such that $$\sum\frac{1}{n_k}<\infty$$ Prove that $\limsup_{k\to\infty} (n_{k+1}-n_k) = \infty$.

Here is what I have so far: Suppose toward a contradiction that $\limsup_{k\to\infty} (n_{k+1}-n_k) = M < \infty$. Then, the sequence {${n_{k+1}-n_k}$} is bounded by $M$. I want to show that this implies $$\sum\frac{1}{n_k}=\infty,$$ a contradiction.

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    what did you try? Hint, assume that $\limsup_{k\to\infty} (n_{k+1}-n_k) = s < \infty$. Prove that $\sum1/n_k = \infty$. (It would more or less end like the harmonic series, times some constant.) – 2017-02-22
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    @Mirko I have edited the question – 2017-02-22
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    try to find a proof for a specific easy case first(and that may give you an idea how to treat the general case). Say $n_1=5,n_2=7,n_3=9$, and in general $n_{k+1}-n_k=2$ for all $k$. How would you prove that $\sum_{k=1}^\infty\frac1{2k+3}=\infty$? That is, try to first prove that $\frac15+\frac17+\frac19+\dots=\infty$. – 2017-02-22
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    No, you can't say that the sequence is bounded by $M.$ But you can say that $n_{k+1}-n_k < M+1$ for all but finitely many $k.$ – 2017-02-22
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    @zhw. And given we work with integers, that actually says that $n_{k+1}-n_k\le M$ for all but finitely many $k$. – 2017-02-22

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You're most of the way there. If $n_{k+1}-n_k\leq M+1$ for large enough $M$ (a more precise application of the $\limsup$ condition), then $n_{k+m}-n_k\leq m(M+1)$, so that $\frac{1}{n_{k+m}}\geq\frac{1}{n_k+m(M+1)}$ for large enough $k$. You can finish by a smart application of the comparison test, see that we have shown that the series $n_k$ is something like $\{\frac{1}{a+bk}\}$.

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    well, $n_{k+1}-n_k\leq M$ – 2017-02-22