6
$\begingroup$

I would like to get a hint on how to establish the convergence of the following sequence:

$$a_{n+1}= a_n + \frac{\sqrt{\vert a_n \vert }}{n^2}$$

where $a_1$ is arbitrary. This is an increasing sequence, so if I could show that it was bounded above I would be done. I cannot figure out how to do that. Any help would be appreciated.

$\endgroup$
5
$\begingroup$

We have $a_n=\sum_{j=1}^{n-1}\frac{\sqrt{|a_j|}}{j^2}+a_1$. Let $M$ such that $M>|a_1|+\frac{\pi^2}6\sqrt M$. Assume that $|a_j|\leqslant M$ for all $1\leqslant j\leqslant n-1$. Then $$|a_n|\leqslant \sqrt M\sum_{j=1}^{n-1}\frac 1{j^2}+|a_1|<M.$$ As $|a_1|<M$, we are done.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

If $a_1\geq 1$, you have that $\sqrt{a_n}<a_n$ for all $n$. Hence, your sequence $(a_n)$ is dominated by the sequence $(b_n)$, where $$ b_{n+1}=b_n\left(1+\frac{1}{n^2}\right),\quad b_1=a_1. $$

Can you show that $(b_n)$ is bounded?

If $a_1<1$ then either the sequence $(a_n)$ is bounded by $1$, or there exists an index $N$ such that $a_N>1$, and the same argument applies as before.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Although not as neat as Davide Giraudo's answer, you may directly apply some convergence test to establish the result.

By redefining $a_1\leftarrow|a_1|$, we may assume that every $a_n$ is nonnegative. Convergence is trivial when $a_1=0$ or $a_n\le1$ for every $n$. So, assume that $a_n>1$ when $n$ is large. Let $b_1=a_1$ and $b_{n+1} = a_{n+1}-a_n$. We want to show that the series $\sum_n b_n$ converges. Now for sufficiently large $n$ and sufficiently small $c>0$, we have \begin{align} \frac{b_{n+1}}{b_n} &= \frac{a_{n+1}-a_n}{a_n-a_{n-1}}\\ &=\frac{(n-1)^2}{n^2}\sqrt{\frac{a_n}{a_{n-1}}}\\ &=\frac{(n-1)^2}{n^2}\sqrt{1+\frac{1}{\sqrt{a_{n-1}}(n-1)^2}}\tag{1}\\ &<\frac{(n-1)^2}{n^2}\sqrt{1+\frac{1}{(n-1)^2}}\\ &=\frac{(n-1)}{n^2}\sqrt{(n-1)^2+1}\\ &<\frac{(n-1)}{n^2}(n-1+c). \end{align} Therefore $n\left(\frac{b_{n+1}}{b_n}-1\right)<-(2-c)+\frac{1+c}{n}<-1$. As equality (1) implies that $\frac{b_{n+1}}{b_n}\rightarrow1$ when $n\rightarrow\infty$, by Raabe's test, $\sum_n b_n$ converges.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.