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Suppose {$a_n$} be a sequence of posetive real numbers such that $a_n\ge a_{n+1}$ for all $n\ge1$ and $\sum_{n=1}^{\infty}a_n <\infty$ then which of the following(s) is true:

A. $\lim_{n\to \infty}a_n=0$

B.$\lim_{n\to \infty}na_n=0$

C.$\lim_{n\to \infty}n^2a_n=0$

D.$lim_{n\to\infty}$$\sum_{m=n}^{\infty}a_m=0$

As the corresponding series convergent so nth term of the series goes to 0. So A is correct.If $a_n=1/n^2$ then C can be removed. I can not understand about other options. Please help.

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  • $\begingroup$ Are there multiple right answers, or is there a single correct answer? $\endgroup$ – Demetri Pananos Feb 27 '15 at 7:50
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    $\begingroup$ @DemetriP: Several of them are true. $\endgroup$ – Regret Feb 27 '15 at 7:51
  • $\begingroup$ C) is false, but your reasoning can be stronger. Try using the ratio test. I'm still thinking about the others. $\endgroup$ – Demetri Pananos Feb 27 '15 at 7:57
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B and D are also true. Let $s_n$ be the sequence of partial sums of $\sum_{n = 1}^\infty a_n$. Then

$$(1)\qquad s_{2n} - s_n = a_{n+1} + \cdots + a_{2n} \ge na_{2n}$$

and $$(2) \qquad s_{2n+1} - s_n = a_{n+1} + \cdots + a_{2n+1} \ge (2n+1)a_{2n+1}.$$

Since $s_n$ converges, $(1)$ implies $2na_{2n} \to 0$, and $(2)$ implies $(2n+1)a_{2n+1} \to 0$. Therefore $na_n \to 0$ and B holds.

Since $\sum_{n = 1}^\infty a_n$, given $\epsilon > 0$, there exists a positive integer $N$ such that $|A - s_n| < \epsilon$ for all $n \ge N$. This implies $$\left|\sum_{m = n}^\infty a_m\right| < \epsilon$$ for all $n > N$. Since $\epsilon$ was arbitary, D follows.

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  • A. True

Classicaly, if $\sum_{n}a_n$ converges, then $a_n\to 0$

  • B. True

Let $n\in \mathbb N$. Then $2na_{2n}\leq 2(a_{2n}+\ldots+a_{n+1})=S_{2n}-S_n$ where $S_n$ the n-th partial sum.

Since $\sum_{n} a_n$ converges, $\lim_{n\to\infty} S_{2n}-S_n =0$.

Hence $\lim_{n\to\infty}2na_{2n}=0$

Moreover $(2n+1)a_{2n+1}=2na_{2n+1}+a{2n+1}\leq 2na_{2n}+a_{2n+1}$

Hence $\lim_{n\to\infty}(2n+1)a_{2n+1}=0$

Hence $\lim_{n\to\infty} na_n=0$

  • D. True

Classically, writing $\sum_{m=n}^\infty a_m = \sum_{m=1}^\infty a_m - \sum_{m=1}^{n-1}a_m$ and recalling that, as a sequence, $\lim_{n\to\infty}\sum_{m=1}^{n-1}a_m=\sum_{m=1}^\infty a_m$ yields the result.

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