Show that the sequence $a_n = a_{n-1}(a_{n-1} + \frac{1}{n})$ is unbounded when $a_1 = 1$

Question

Show that the sequence $a_n = a_{n-1}(a_{n-1} + \frac{1}{n})$ is unbounded when $a_1 = 1$

I know intuitively that it is unbounded but how can I show that it is unbounded. In particular I want to show that it is monotone increasing and then show there cannot be an upper bound. However I can't seem to figure out what I should do to show that it is monotone increasing.

Answer 1

It's clear that $a_i>1 \implies a_n>a_{n-1}+\frac1n \implies a_i>1+\frac12+\frac13+\dotsb+\frac1i$ but clearly the RHS is not bounded and we're done

Answer 2

Alternatively, first we show $a_n > 1, \forall n \ge 2$ by induction. $a_2 = 1\cdot(1+\frac{1}{2})= \frac{3}{2} > 1$. Assume $a_n > 1, n \ge 2 \implies a_{n+1} =a_n^2++\dfrac{a_n}{n+1}> 1+\dfrac{1}{n+1} > 1\implies a_{n+1} > 1\implies a_n > 1, \forall n \ge 2$. Thus using the recursive formula we have: $a_n = a_{n-1}^2+\dfrac{a_{n-1}}{n}> a_{n-1}^2 \implies a_n > a_{n-1}^2$. Apply this new inequality repeatedly on the way to the $2$nd term: $a_n > a_{n-1}^2> (a_{n-2})^{2^2}> (a_{n-3})^{2^3}> \cdots> (a_2)^{2^{n-2}}= \left(\frac{3}{2}\right)^{2^{n-2}}$. This shows $a_n$ is unbounded.

Answer 3

If the series has a limit then it must be a stationary point which is

$$a_n=a_{n-1}(a_{n-1}+\frac1{n})|_{n \to \infty}$$

$$a_{\infty}=a_{\infty}^2$$

so it is either $a_{\infty}=0$ or $a_{\infty}=1$.

But for $a_1=1$ it is already $a_2=\frac{3}{2}$ and the series is monotonically increasing, thus it is unbounded. It simply cannot reach either $0$ or $1$.

In order to find the exact value where it becomes unbounded, you need to find the limit of $a_1$ for each $1=a_{n-1}(a_{n-1}+\frac1{n})$ (and similarly $0=a_{n-1}(a_{n-1}+\frac1{n})$). For example find the solution for $a_{1,1}=a_{1},1=a_{1}(a_{1}+\frac1{2})$ then $a_{1,2}=a_{1},1=a_{2}(a_{2}+\frac1{3})$, $a_{1,3}=a_{1},1=a_{3}(a_{3}+\frac1{4})$.

You will get a decreasing series (obviously, right?) $a_{1,m}$ and its limit $a_{1,\infty}$ will be the breaking point bellow which or at which the series will tend to $1$ or $0$, and above which to infinity.

But the solution of $1=a_{1}(a_{1}+\frac1{2})$, $a_1=\frac{\sqrt{17}}{4}-\frac{1}{4} \approx 0.78$ is already less than $1$, proving yet again that your series is unbounded.