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Let $\{a_n\}$ be a sequence of positive terms. Pick out the cases which imply that $\sum a_n$ is convergent.

a. $\displaystyle\lim_{n\to\infty}n^{3/2}a_n = \frac{3}{2}$.

b. $\sum n^2a_n^2 < \infty$.

c. $\displaystyle\frac{a_{n+1}}{a_n} < \left(\frac{n}{n + 1}\right)^2$, for all $n$.

How should I proceed? I am completely stuck on it.

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  • $\begingroup$ Alternative 'a' is an easy yes. $\endgroup$
    – ncmathsadist
    Jan 5, 2013 at 2:31
  • $\begingroup$ For b., let $p=na_n$ and $q=1/n$. Note that $a_n=pq\le{1\over2}(p^2+q^2)={1\over2}(n^2a_n^2+{1\over n^2})$. $\endgroup$
    – David Mitra
    Jan 5, 2013 at 2:58
  • $\begingroup$ For c, observe that $a_i < (\frac {i-1}{i})^2 a_{i-1} < (\frac {1}{i})^2 a_i$. $\endgroup$
    – Calvin Lin
    Jan 5, 2013 at 3:02
  • $\begingroup$ Case $(b)$, you have that $\lim_{n \to \infty } n^2\,a^2_n = 0$. $\endgroup$
    – Mhenni Benghorbal
    Jan 5, 2013 at 4:03
  • $\begingroup$ @MhenniBenghorbal How does this hlep ? Could you please xplain ? $\endgroup$
    – user114873
    May 26, 2017 at 6:50

1 Answer 1

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Outlines:

a) When $n$ is large enough, $n^{3/2}a_n\lt 2$. Now it should be a straightforward comparison with $\displaystyle\sum \frac{2}{n^{3/2}}$.

c) Let $a_1=b$. Then $a_2\lt b\dfrac{1^2}{2^2}=\dfrac{b}{2^2}$. Therefore $a_3\lt b\dfrac{1^2}{2^2}\dfrac{2^2}{3^2}=\dfrac{b}{3^2}$. And so on. But $\displaystyle\sum \frac{1}{n^2}$ converges.

b) I am not happy about my suggestion for this one! Please see the Comment by David Mitra for a better approach that uses much less machinery.

By the Cauchy-Schwarz Inequality, we have, for positive $x_k$, $y_k$, that $$\left(\sum_{1}^n x_k y_k\right)^2\le \left(\sum_1^n x_k^2\right)\left(\sum_1^n y_k^2\right).$$ Let $x_k=ka_k$ and let $y_k=\frac{1}{k}$. Then $$\left(\sum_{1}^n a_k\right)^2\le \left(\sum_1^n k^2a_k^2\right)\left(\sum_1^n \frac{1}{k^2}\right).$$

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  • $\begingroup$ No need for C-S: $a_i=x_iy_i\le{1\over 2}(x_i^2+y_i^2)={1\over2}(i^2a_i^2+{1\over i^2})$. $\endgroup$
    – David Mitra
    Jan 5, 2013 at 3:06
  • $\begingroup$ Thanks. I saw your earlier comment after posting. The OP will see it too, and will learn how to do the problem without the excessive machinery that I used. $\endgroup$
    – André Nicolas
    Jan 5, 2013 at 3:11

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