Given a sequence $\left\{a_n\right\}_{n=0}^\infty$, it is clearly wrong that if $\lim_{n\to\infty}[a_n-a_{n+1}]=0$ then $a_n$ is convergent, but if it is known that $\lim_{n\to\infty}[a_n-a_{2n}]=0$, will then $a_n$ converge? Can anyone find a short proof of that or a counterexample?
Disclaimer: This is not homework, but rather something I forgot from Calculus I.


If $a_n=H_n=\sum_{j=1}^{n}\frac{1}{j}$ and $a_n=\log n$ are counterexamples to the first assumption, $a_n = \log \log n$ is a counterexample to the second one.

  • $\begingroup$ Thanks. I had the first example in mind when I wrote that 'it is clearly wrong that...', but the second example is a very nice one $\endgroup$ – Dennis Gulko Nov 19 '12 at 10:57

We have (a)$\Rightarrow$(b):

(a) If $\lim_{n\to\infty}(a_n-a_{2n})=0$, then $\{a_n\}$ converges.

(b) If $\lim_{n\to\infty}(b_n-b_{n+1})=0$, then $\{b_n\}$ converges.

(Reason: if $\lim_{n\to\infty}(b_n-b_{n+1})=0$, define $a_{2^n}=b_n$ and every other $a_i=0$.) Since you know that (b) is clearly wrong, I guess the conclusion is clear.


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