# Does $\frac{a_{n+1}}{a_n} < (\frac{n}{n+1})^2$ imply $\sum a_n <\infty$ here?

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

1. $\lim n^{3/2}a_n=3/2$

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

3. $\frac{a_{n+1}}{a_n} < (\frac{n}{n+1})^2$, for all $n$.

My Attempt.

1. True. The given condition implies: $\lim_{n\to \infty} \frac{a_n}{1/n^{3/2}}=3/2$. Now $\sum 1/n^{3/2}<\infty$ so does $\sum a_n$.

2. True. Using Holder's inequality we have: $$\sum a_n \le (\sum 1/n^2)^{1/2}~(\sum n^2a_n^2)^{1/2}<\infty$$

3. I think here Ratio test cannot help us...So how can I proceed in this option...? Thank you.

• Which upper bound for $a_n$ in terms of $a_1$ and $n$ can you deduce from 3.? – Did Aug 25 '18 at 7:12
• @Did...Thank you I got it...we actually get..$a_{n+1}<\frac{a_1}{(n+1)^2}$...Right..? – Indrajit Ghosh Aug 25 '18 at 7:15
• Exactly. Well done. – Did Aug 25 '18 at 7:16
• @IndrajitGhosh Why do you think ratio test does not work for 3? – Empty Aug 25 '18 at 7:37
• @Empty...The $\lim$ operation on the both side of the inequality may produce $\le$...so we cannot conclude $\lim \frac{a_{n+1}}{a_n}<1$... – Indrajit Ghosh Aug 25 '18 at 7:40

As regards 3), more generally, if $\{a_n\}_{n\geq 1}$ and $\{b_n\}_{n\geq 1}$ are sequences of positive terms such that $$\frac{a_{n+1}}{a_n}\le \frac{b_{n+1}}{b_n}$$ and $\sum_{n}b_n$ is convergent then also $\sum_{n}a_n$ is convergent. In fact $$a_{n}=a_1\prod_{k=1}^{n-1}\frac{a_{k+1}}{a_k}\leq a_1 \prod_{k=1}^{n-1}\frac{b_{k+1}}{b_k}\leq \frac{a_{1}}{b_1}\, b_n$$ and convergence follows by the comparison test. In your case $b_n=1/n^2$.

• Nice one....... – Indrajit Ghosh Aug 25 '18 at 7:24

As an alternative to the elegant way found in the comments, by Raabe’s test we have

$$n\left(\frac{a_{n}}{a_{n+1}}-1\right)=n\frac{2n+1}{n^2} \to 2>1$$

Since $$\frac{a_{n+1}}{a_n} < \left(\frac{n}{n+1}\right)^2,$$

hence $$a_n=a_1\cdot\frac{a_2}{a_1}\cdot\frac{a_3}{a_2}\cdots\frac{a_n}{a_{n-1}}<a_1\cdot\left(\frac{1}{2}\right)^2\cdot\left(\frac{2}{3}\right)^2\cdots\left(\frac{n-1}{n}\right)^2=\frac{a_1}{n^2}.$$

Thus $$\sum^\infty a_n\leq\sum^\infty\frac{a_1}{n^2}=a_1\cdot\sum^\infty\frac{1}{n^2}=\frac{\pi^2a_1}{6}<\infty,$$which is convergent.