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Let $a_n > 0$ and for all $n$ let $$\sum\limits_{j=n}^{2n} a_j \le \dfrac 1n $$ Prove or give a counterexample to the statement $$\sum\limits_{j=1}^{\infty} a_j < \infty$$

Not sure where to start, a push in the right direction would be great. Thanks!

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up vote 3 down vote accepted

Think about what bounds you can put on these:

$$b_k = \sum\limits_{j=2^k}^{2^{k+1}-1} a_j$$

and note that $\sum\limits_1^\infty a_n = \sum\limits_0^\infty b_k$.

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So $b_0 \le 1$, $b_1 \le 1/2$, $b_2 \le 1/4$, $b_3 \le 1/8$, and in general $b_k \le 1/2^k$. Therefore $b_0 + b_1 + b_2 + ... + b_k + ... \le 1 + 1/2 + 1/4 + ... + 1/2^k + ... $ Since $\sum_1^{\infty} a_n = \sum_0^{\infty} b_k$ then we have that $\sum_1^{\infty} a_n = \sum_0^{\infty} b_k \le \sum_0^{\infty} 1/2^k$ which is a convergent geometric series and therefore we have convergence of $\sum a_n$ and its finite? Is this correct? – user52709 Dec 12 '12 at 1:24
I think so, yes. – Itai Zukerman Dec 12 '12 at 1:38

Consider sum of sums $\sum_{i=1}^\infty \sum_{k=i}^{2k} a_j$.

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Well, my formulation is not perfectly correct, but the idea should be evident. – Valtteri Dec 12 '12 at 0:34

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