# Limit of $a_{n+1} = a_{n}(1+1/n^2)$

I read the following exercise :

Let $$(a_n)$$ be a sequence such as :

$$a_1 = 2$$, $$a_{n+1} = a_n.(1+1/n^2)$$

Prove that $$\lim_{n \rightarrow \infty} a_n$$ exists.

Hint : $$\ln : \: (0, \infty) \rightarrow \mathbb{R} \:$$ is a continuous, increasing function such as $$\ln(1+x) \leq x$$ and $$\ln(ab) = \ln(a)+\ln(b)$$.

It is quite clear that $$a_n$$ is increasing but then I don't really know how to proceed. Could you please help me ?

Thanks.

• write $a_{n+1}$ by substituting the value of a(n).... recursively... then, use rienmann sums Nov 22, 2020 at 15:37
• @PNDas Yes I agree but how ? And what would the ln function be useful for ? Nov 22, 2020 at 15:37
• If you want to use $\ln$ then see $\ln(a_{n+1})=\ln(a_n)+\ln(1+\frac 1{n^2})\leq\ln(a_n)+\frac 1{n^2}\leq a_{n-1}+\frac 1{{n-1}^2}+\frac 1{n^2}$, inductively, $\ln(a_{n+1})\leq a_1+ \sum_{m=1}^n\frac 1{m^2}\leq a_1+ \sum_{m=1}^{\infty}\frac 1{m^2}= a_1+ \frac {{\pi}^2}6$ Nov 22, 2020 at 15:45
• @PNDas How does the limit of the ratio being 1 show above boundedness? Take $a_n=n^2$ for example. Limit of the ratio is still 1 but the sequence is unbounded above. Really not following the logic. Nov 22, 2020 at 18:44
• @CogitoErgoCogitoSum,yeah you are right, but I didn't say limit of ratio =1 then sequence is bounded. Nov 23, 2020 at 5:50

As $$\ln x$$ is increasing, $$b_n = \ln a_n$$ is an increasing sequence.

Also $$b_{n+1}= \sum_{k=1}^n \ln(1+1/k^2) \le \sum_{k=1}^n 1/k^2$$

according to the provided hint.

As the series on the RHS is convergent, the sequence $$\{b_n\}$$ is bounded. Being also increasing, it is convergent. Therefore so is $$\{a_n\}$$.

Prove by induction that $$\ln a_n \le \sum_{k=1}^{n-1}\frac1{k^2}$$ for all $$n \ge 2$$.

For $$n=2$$ we have $$\ln a_2 = \ln \frac52 < 1$$ and if we assume that $$\ln a_n \le \sum_{k=1}^{n-1}\frac1{k^2}$$ for some $$n$$ then $$\ln a_{n+1} = \ln a_n + \ln\left(1+\frac1{n^2}\right) \le \sum_{k=1}^{n-1}\frac1{k^2} + \frac1{n^2} = \sum_{k=1}^n \frac1{k^2}$$ which completes the proof.

Hence $$\ln a_{n} \le \sum_{k=1}^\infty \frac1{k^2} = \frac{\pi^2}6 \implies a_n \le e^{\frac{\pi^2}6}$$ so your sequence is bounded.

hint

By induction, it is easy to see that $$(\forall n\in \Bbb N) \;\; a_n>0$$ put $$b_n=\ln(a_n)$$

then

$$b_{n+1}=b_n+\ln(1+\frac{1}{n^2})$$ $$\le b_n +\frac{1}{n^2}$$

Take $$\ln$$ about both sides, we have $$\ln(a_{n+1})=\ln(a_n)+\ln(1+\frac{1}{n^2})\leq\ln(a_n)+\frac{1}{n^2}$$.

Then $$\ln(a_{n+1})-\ln(a_n)\leq\frac{1}{n^2}$$.

adding them above $$n=1,2,\cdots$$ $$\underset{n\to\infty}{\lim\sup}\ln(a_n)-\ln(a_1)\leq\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$ So,$$\underset{n\to\infty}{\lim\sup}\ln(a_n)\leq \ln(a_1)+\frac{\pi^2}{6}$$, and $$\underset{n\to\infty}{\lim\sup} a_n\leq 2e^{\frac{\pi^2}{6}}$$. Therefore, the sequence is bounded. Furthermore, the sequence has a limit point due to increasing.

• I think it is more interesting to evaluate the limit point. Nov 22, 2020 at 15:58

By using $$\ln (1+x)\le x$$ for $$x\ge 0$$, we have $$1+{1\over n^2}\le e^{1\over n^2}$$ hence $$a_n$$ is bounded above by another sequence $$b_n$$ where $$b_{n+1}=b_ne^{1\over n^2}\quad,\quad b_1=4$$ Since $$b_n\to b_1e^{\sum_{i=1}^\infty {1\over i^2}}=4e^{\pi^2\over 6}$$ then $$a_n$$ also tends to a limit because it is strictly increasing and bounded from above.