# How do I prove that there exists a such subsequence?

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 $$\sum_k^\infty {a_{n_k}}<\infty$$?

• @PeterForeman I just edited my post! I missed a condition Nov 20 '19 at 16:51

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

• You want a bit more, namely, that $n_k<n_{k+1}$ for all $k$ so you truly have a subsequence. Nov 20 '19 at 17:33

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.

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 $$a_{n_k} < \frac 1{2^k}$$.