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Let $\{a_n\}$ be a sequence of positive number such that $$a_1>a_2>a_3>...$$

then which of the followings are true ?

  1. $\displaystyle\lim_{n\to\infty}a_n=0$
  2. $\displaystyle\lim_{n\to\infty}\frac{a_n}{n}=0$
  3. $\displaystyle\sum_{n=1}^{n=\infty}\frac{a_n}{n}$ converges
  4. $\displaystyle\sum_{n=1}^{n=\infty}\frac{a_n}{n^2}$ converges

For (1) and (3) , $a_n=2+\frac{1}{n}$ and $a_n=\frac{1}{log n}$.So 1st is false and 3 is false due to 'Cauchy condensation test' .But I am unable to prove or disprove 2 and 4.Please help.

Thanks in advance.

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    $\begingroup$ $(2)$ and $(4)$ is true. Hint: the sequence $a_i$ is bounded from above by $a_1$. $\endgroup$ – achille hui Jun 11 '16 at 5:34
  • $\begingroup$ For (2), we have $0\le \frac{a_n}{n}\le \frac{a_1}{n}$ so the limit is $0$ by squeezing. $\endgroup$ – André Nicolas Jun 11 '16 at 5:35
  • $\begingroup$ To disrepute (1), consider the counter-example $a_n = 1 + \frac{1}{n}$. Furthermore, this sequence may be used to disrepute (3) because $s_k = \sum_{n = 1}^{k} \frac{a_n}{n} > \sum_{n = 1}^{k} \frac{1}{n}$ which diverges as $k \rightarrow \infty$. $\endgroup$ – Steven Harding Jun 11 '16 at 5:41
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hint: Consider (as Andre Nicolas pointed out) $0\leq \frac{a_n}{n}\leq \frac{a_1}{n}$, with squeeze theorem for $2$.

For $4$, as $a_n$ is decreasing sequence so consider $\sum\frac{a_n}{n^2}<\sum\frac{a_1}{n^2}=a_1\sum\frac{1}{n^2}$, use comparison test.

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  • $\begingroup$ I can't undestand,please explain a little more $\endgroup$ – Bhusan Jun 11 '16 at 5:33
  • $\begingroup$ But for 2 I needed a general proof $\endgroup$ – Bhusan Jun 11 '16 at 5:36
  • $\begingroup$ @Bhusan I have edited the answer. $\endgroup$ – Kushal Bhuyan Jun 11 '16 at 5:45
  • $\begingroup$ Notice that (from the MCT) $\lim_{n \rightarrow \infty} a_n = L \geq 0$. Let $\epsilon > 0$. There exists $N_1 \in \mathbb{N}$ such that $|a_n - L| < \dfrac{\epsilon}{2}$ for $n \geq N_1$. There also exists $N_2 \in \mathbb{N}$ such that $\dfrac{L}{n} < \frac{\epsilon}{2}$ for $n \geq N_2$. Then, for $n \geq \max\{N_1,N_2\}$, we have $|\dfrac{a_n}{n}| = \dfrac{|a_n|}{n} = \dfrac{|a_n - L + L|}{n} \leq \dfrac{|a_n - L| + |L|}{n} < \dfrac{\epsilon/2 + L}{n} \leq \epsilon/2 + L/n < \epsilon/2 + \epsilon/2 = \epsilon$. You can eliminate a lot of these absolute value signs. $\endgroup$ – Steven Harding Jun 11 '16 at 5:55
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$0<\frac{a_n}{n}<\frac{a_1}{n} \rightarrow 0$, so Squeeze Theorem,

and

$\sum \frac{a_n}{n^2} < \sum \frac{a_1}{n^2}=a_1\pi^2/6$

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