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Let $a_n$ be a sequence of nonnegative reals such that $a_{n+1} \leq a_n$ and $\lim_{n\to \infty} a_n = 0$

Then, how do I prove that there exists a subsequence such that $\sum_k^\infty {a_{n_k}}<\infty$?

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  • $\begingroup$ @PeterForeman I just edited my post! I missed a condition $\endgroup$
    – user11
    Nov 20 '19 at 16:51
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For each $k\in\mathbb{N}^+$ choose $a_{n_k}$ such that $a_{n_k} < \frac{1}{k^2}$. This is possible since $a_n$ is a decreasing sequence of nonnegative reals with limit $0$. Then $\sum_{k = 1}^\infty a_{n_k}$ converges by comparison to $\sum_{k = 1}^\infty \frac{1}{k^2}$.

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  • $\begingroup$ You want a bit more, namely, that $n_k<n_{k+1}$ for all $k$ so you truly have a subsequence. $\endgroup$ Nov 20 '19 at 17:33
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The Direct Comparison test states that the infinite series $\sum b_{n}$ converges and $0\leq a_{n}\leq b_{n} $ for all sufficiently large $n$ (that is, for all $n>N$ for some fixed value N), then the infinite series $\sum a_{n}$ also converges.

Then since you know that $a_n\rightarrow 0$ (monotonically which implies that are all positive) this means that $\forall \epsilon>0,\exists N(\epsilon)$ such that for all $n\geq N(\epsilon)$ $$a_n<\epsilon$$

Take for instance $\epsilon=\frac{1}{n^2}$ for any $n\in\mathbb N\backslash\{0\}$, then there will be an $N(\epsilon)$ such that whenever $n_k\geq N(\epsilon)$ $$a_{n_k}<\epsilon=\frac{1}{n^2}$$

But you know that the series $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges, and then applying the direct comparison test you conclude that $\sum_{k=1}^\infty a_{n_k}$ converges as well.

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Well, you can use a flyswatter approach.

$\sum_{k=1}^\infty \frac 1{2^k}= 1$.

So if you take a subsequence $a_{n_k}$ so that each $a_{n_k}: 0 \le a_{n_k} < \frac 1{2^k}$ you'd have $\sum_{k=1}^\infty a_{n_k} < \sum_{k=1}^\infty \frac 1{2^k}= 1$.

And since $a_{n}\to 0$ for every $\frac 1{2^k}$ there is an $N_k$ so that $n > N_k\implies |a_n|=a_n < \frac 1{2^k}$. So inductively choose each $n_k$ so that $n_k > n_{k-1}$ and $a_{n_k} < \frac 1{2^k}$.

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