# Prove that if $a_1 + a_2 + \ldots$ converges then $a_1+2a_2+4a_4+8 a_8+\ldots$ converges and $\lim na_n=0$

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.

(a)

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?

(b)

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*}

• Note that condition two is called Cauchy condensation.
– user296602
Commented Dec 22, 2015 at 8:36
• 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. Commented Dec 22, 2015 at 10:19
• @gebruiker Edited. Commented Dec 22, 2015 at 10:29
• @luka5z You've taken away the fun out of it by solving it yourself, haha...
– user98186
Commented Dec 22, 2015 at 15:42