# 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 Dec 22 '15 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. – gebruiker Dec 22 '15 at 10:19
• @gebruiker Edited. – luka5z Dec 22 '15 at 10:29
• @luka5z You've taken away the fun out of it by solving it yourself, haha... – user98186 Dec 22 '15 at 15:42