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Consider the sequence $\{a_n\}$ defined by $$a_n=\frac1{(n+1)^{3/2}}+\dots+\frac1{(2n)^{3/2}}\;.$$ As $n\to\infty$, the sequence $a_n$

$\quad$(A) converges to $0$.
$\quad$(B) diverges to $\infty$.
$\quad$(C) is bounded but does not converge.
$\quad$(D) convertes to a positive number.

Please help to solve this problem.

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2 Answers 2

up vote 1 down vote accepted

HINT: The series $\displaystyle\sum_{n\ge 1}\frac1{n^{3/2}}$ converges. (Why?) Therefore the tail sums $t_n=\displaystyle\sum_{k\ge n}\frac1{k^{3/2}}$ converge to $0$. (Why?) How is $a_n$ related to $t_{n+1}$?

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The series $\displaystyle\sum_{n\ge 1}\frac1{n^{3/2}}$ converges as we know The series $\displaystyle\sum_{n\ge 1}\frac1{n^{p}}$ converges for $p>1$ and diverges for $p \leq 1.$So by comparison test ,we can say the series will converge.So the answer is between $(A)$ and $(D)$. – user53386 Jan 19 '13 at 19:49
@user53386: But you can use what I said about the tail sums to decide between (A) and (D). – Brian M. Scott Jan 19 '13 at 19:51
Thanks a lot sir.I have got it. option $(A)$ will be the correct one. – user53386 Jan 19 '13 at 19:57
@user53386: That’s right. And you’re very welcome. – Brian M. Scott Jan 19 '13 at 19:57

$a_n \leq n \times \dfrac{1}{n^{3/2}} $

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