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$\{xn\}$ 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.

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  • $\begingroup$ That's four questions. Anyway, my guesses are yes, no, yes, no. $\endgroup$ – Angina Seng Dec 8 '19 at 3:44
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  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.
  2. False. Answered elsewhere.
  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 $n \geq 4$.)
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  • $\begingroup$ I'm sceptical about 2. For a start, why does $(nx_n)$ converge? $\endgroup$ – Angina Seng Dec 8 '19 at 3:55
  • $\begingroup$ For 2., see user284331's reply. $\endgroup$ – Angina Seng Dec 8 '19 at 4:00
  • $\begingroup$ I can't make sense of user284331's reply since it is currently variable salad. $\endgroup$ – Eric Towers Dec 8 '19 at 4:02
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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.

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

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