Let $a_1,a_2,a_3,\ldots$ be a decreasing sequence of positive numbers. Show that

(a) if $a_1+a_2+\ldots$ converges then $\lim_{n\rightarrow\infty} n a_n=0$

(b) $a_1+a_2+\ldots$ converges if and only if $a_1+2 a_2+4 a_4 +\ldots $ converges.


If $\sum a_i$ converges then for any $\epsilon>0$ there is natural number $N_1$ such that if $n>N_1$ then $$2n \cdot a_{2n} \le\sum_{i=n}^{2n} a_i <\epsilon$$ We cam deal in the same way with the odd terms and for given $\epsilon>0$ find $N_2$ such that $$(2n+1) \cdot a_{2n+1} \le\sum_{i=n+1}^{2n+1} a_i <\epsilon$$ So for every $\epsilon>0$ there is $N=\max\{N_1,N_2\}$ such that whenever $n>N$ then $na_n <\epsilon$.

Is this the correct way of proving that fascinating fact?


If the second series converges then since $a_1,a_2,\ldots$ is decreasing sequence of nonegative numbers, from comparison test we know that the first series converges too.

For the converse I will show that partial sums of the second series are bounded. $$\begin{align*} a_1+\frac12\sum_{i=1}^N2^ia_{2^i}&=a_1+a_2+2a_4+4a_8+\dots+2^{N-1}a_{2^N}\\ &\leq a_1+ a_2+a_3+a_4+a_5+a_6+a_7+a_8+\dots+a_{2^{N-1}+1}+\dots+a_{2^N-1}+a_{2^N}\\ &\leq \sum_{i=1}^\infty a_i<\infty \end{align*}$$

  • 3
    $\begingroup$ Note that condition two is called Cauchy condensation. $\endgroup$
    – user296602
    Dec 22, 2015 at 8:36
  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures cannot be searched and are inaccessible to those using screen readers. Please edit your question accordingly. $\endgroup$
    – gebruiker
    Dec 22, 2015 at 10:19
  • $\begingroup$ @gebruiker Edited. $\endgroup$
    – luka5z
    Dec 22, 2015 at 10:29
  • 2
    $\begingroup$ @luka5z You've taken away the fun out of it by solving it yourself, haha... $\endgroup$
    – user98186
    Dec 22, 2015 at 15:42

1 Answer 1


If you're interested, the Cauchy condensation test is actually a special case of the Schlömilch test. The proof is actually relatively straightforward and very simply stated (with some missing details) here: http://arxiv.org/pdf/1011.4697.pdf.

The generalization covers many of the same series that the integral test would otherwise cover, so its utility is partially diminished.

  • $\begingroup$ Nice, Ill check this out. $\endgroup$
    – luka5z
    Dec 22, 2015 at 18:14
  • 1
    $\begingroup$ Yes! It's very interesting. The relevant part is just that s: N->N is monotonically increasing, and that it doesn't grow "too fast," since it is acting as a "weight" and would not always converge if the function grew to quickly. $\endgroup$ Dec 22, 2015 at 18:29

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