0
$\begingroup$

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.

$\endgroup$

2 Answers 2

Reset to default
6
$\begingroup$

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

$\endgroup$
0
1
$\begingroup$

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 $t_{n+1}$?

$\endgroup$
4
  • $\begingroup$ 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 $(D)$. $\endgroup$
    – user53386
    Jan 19, 2013 at 19:49
  • $\begingroup$ @user53386: But you can use what I said about the tail sums to decide between (A) and (D). $\endgroup$
    – Brian M. Scott
    Jan 19, 2013 at 19:51
  • $\begingroup$ Thanks a lot sir.I have got it. option $(A)$ will be the correct one. $\endgroup$
    – user53386
    Jan 19, 2013 at 19:57
  • $\begingroup$ @user53386: That’s right. And you’re very welcome. $\endgroup$
    – Brian M. Scott
    Jan 19, 2013 at 19:57

You must log in to answer this question.