# Multiple choice question on sequence and series

$$\{xn\}$$ be a sequence of positive real numbers such that $$\sum_{n=1}^{\infty} x_n$$ converges. which of following are true

1. The series $$\sum_{n=1}^{\infty} \sqrt{x_nx_{n+1}}$$ converges
2. $$lim_{n\rightarrow \infty}nx_n =0$$
3. The series $$\sum_{n=1}^{\infty} sin^2 x_n$$ converges
4. $$\sum_{n=1}^{\infty} \frac{\sqrt{x_n}}{1+\sqrt{x_n}}$$ converges

I think first can be proved by limit comparison test. For two by divergence test, $$lim_{n\rightarrow \infty}x_n =0$$, but counter example for the given statement I dont have. 4 by direct comparison convergent.

• That's four questions. Anyway, my guesses are yes, no, yes, no. – Angina Seng Dec 8 '19 at 3:44

1. True. By the AM-GM, $$\sqrt{x_n x_{n+1}} \leq \frac{x_n + x_{n+1}}{2}$$, so \begin{align*} \sum_{n=1}^\infty \sqrt{x_n x_{n+1}} &\leq \sum_{n=1}^\infty \frac{x_n + x_{n+1}}{2} \\ &= \frac{1}{2} \sum_{n=1}^\infty x_n + \frac{1}{2} \sum_{n=2}^\infty x_n \text{,} \end{align*} so converges.
3. True. Since $$\sum x_n$$ converges, $$x_n \rightarrow 0$$. So there is an $$N > 0$$ such that for all $$n > N$$, $$x_n < 1$$. Then, for $$n > N$$,
$$0 \leq \sin^2 x_n \leq x_n^2 < x_n$$ (... because $$\sin'(x) \leq 1$$ and $$\sin''(x) < 0$$ for $$x \in [0,1]$$). By the comparison test, since $$\sum_{n=1}^\infty x_n$$ converges, so does $$\sum_{n=1}^\infty \sin^2 x_n$$.
4. False. Consider $$x_n = n^{-3/2}$$. Then $$\frac{\sqrt{x_n}}{1+\sqrt{x_n}} = \frac{n^{-3/4}}{1+n^{-3/4}} \cdot \frac{n^{3/4}}{n^{3/4}} = \frac{1}{n^{3/4}+1} \text{,}$$ which is greater than $$n^{-1}$$ for sufficiently large $$n$$. (In fact, for $$n \geq 4$$.)
• I'm sceptical about 2. For a start, why does $(nx_n)$ converge? – Angina Seng Dec 8 '19 at 3:55
• For 2., see user284331's reply. – Angina Seng Dec 8 '19 at 4:00
• I can't make sense of user284331's reply since it is currently variable salad. – Eric Towers Dec 8 '19 at 4:02

1) $$\displaystyle\sum\sqrt{x_{n}x_{n+1}}\leq\left(\sum x_{n}\right)^{1/2}\left(\sum x_{n+1}\right)^{1/2}$$.

2) Consider $$a_{n}=1/k^{2}$$ for $$n=k^{4}$$ and $$a_{n}=1/n^{2}$$ otherwise, $$(na_{n})$$ has no limit.

3) $$\sin^{2}x_{n}=\dfrac{\sin^{2}x_{n}}{x_{n}^{2}}x_{n}^{2}\leq x_{n}^{2}\leq x_{n}$$ for $$n$$ large.

• For the case 2 you need to define $x_n$ a bit differently since they are supposed to be positive. – trancelocation Dec 8 '19 at 4:04