# Show that the sequence $a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$ is upper bounded

Let $\{a_{n}\}$ be defined with $a_{1}\in(0,1)$, and $$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$$ for all $n\gt 0$. Show that the sequence is upper bounded.

My idea: since $$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$ then $$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}$$ then $$\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}$$ so $$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}$$ since $$a_{n+1}>a_{n}\Longrightarrow \dfrac{1}{a_{i}+i^2}<\dfrac{1}{a_{1}+i^2}$$ so $$\dfrac{1}{a_{n+1}}>\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)$$ But the RHS might be $\lt0$ for a sufficiently large starting value; for instance, with $a_{1}=\dfrac{99}{100}$ then $$\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)<0,n\to\infty$$

see:

so this method won't let me bound the series and I don't know what else to do.

• since $a_1\in (0,1)$ your sequence is obviously bounded from below by, e.g. $0$. I don't think there's an upper bound. – dinosaur Feb 18 '14 at 15:44
• Have you looked at the trend of values of the sequence for various starting values? Since it seems highly likely that the upper bound will go to $\infty$ as $a_1\to 1$, being able to conjecture a form for the upper bound in terms of $a_1$ (e.g., that the sequence is bounded by $(1-a_1)^{-1}$ or something similar) might help with a proof. – Steven Stadnicki Feb 18 '14 at 16:28
• My idea: we must prove this $a_{n}>f(a_{1})?$ – math110 Feb 18 '14 at 16:53
• @math110 Actually, I think you want to prove the opposite: that $a_n\lt f(a_1)$ for some appropriate $f$. Preliminary browsing around Alpha suggests that $a_n$ may in fact (slowly) converge to something like $1+(1-a_1)^{-1}$, so that might be an angle to start with. – Steven Stadnicki Feb 18 '14 at 17:32

Here is a solution: Since $(a_{n})$ is monotone increasing, either it remains bounded or it diverges to $+\infty$. Assume that $a_{n} \nearrow +\infty$.

Observation 1. Referring to OP's identity

$$\frac{1}{a_{n}} - \frac{1}{a_{n+1}} = \frac{1}{a_{n} + n^{2}},$$

we have

$$\frac{1}{a_{n}} - \frac{1}{a_{m+1}} = \sum_{k=n}^{m} \frac{1}{a_{k} + k^{2}}.$$

Taking $m \to \infty$, it follows that

\begin{align*} \frac{1}{a_{n}} = \sum_{k=n}^{\infty} \frac{1}{a_{k} + k^{2}} \leq \sum_{k=n}^{\infty} \frac{1}{k^{2} - k} = \frac{1}{n-1}. \end{align*}

Thus we must have $n-1 \leq a_{n}$ for any $n \geq 2$.

Observation 2. Now we prove that $a_{n} \leq a_{1} n$ for any $n$. Indeed, this is trivial when $n = 1$. Also, if it holds for $n$ then

$$a_{n+1} = a_{n} \left( 1 + \frac{a_{n}}{n^{2}} \right) \leq n a_{1} \left( 1 + \frac{1}{n} \right) = (n+1)a_{1}.$$

Therefore by induction, we have the desired estimate.

Conclusion. Combining two observation, it follows that

$$n - 1 \leq a_{n} \leq na_{1} \quad \Longrightarrow \quad 1 - \frac{1}{n} \leq a_{1}.$$

for $n \geq 2$. Taking $n\to\infty$, we have $1 \leq a_{1}$, a contradiction! Therefore $(a_{n})$ must remain bounded.

Addendum. Now let us investigate an asymptotic behavior of the limit $f(a_{1}) = \lim_{n\to\infty} a_{n}$. We have

\begin{align*} \frac{1}{a_{1}} - \frac{1}{f(a_{1})} &= \sum_{n=1}^{\infty} \frac{1}{n^{2} + a_{n}} \geq \sum_{n=1}^{\infty} \frac{1}{n^{2} + n} = 1, \end{align*}

So we have a lower bound.

$$f(a_{1}) \geq \frac{a_{1}}{1 - a_{1}}.$$

• I think the last line should be $1\leq a_2$ ($n\geq 2$). And $1\leq a_2$ is no contradiction. – pisoir Feb 18 '14 at 17:44
• @pisoir, have you read Observation 2 carefully? – Sangchul Lee Feb 18 '14 at 17:51
• Very nice. +1.  – Did Feb 18 '14 at 17:58
• @sos440 Maybe I am missing something, but in Obser.2 you showed that e.g. $a_2\leq 2a_1$. And conclusion is that e.g. $1\leq a_2\leq 2a_1$. For $n=1$, you showed $0\leq a_1 \leq a_1$. I'm sorry I don't see any contradiction. – pisoir Feb 18 '14 at 17:59
• @pisoir, the contradiction occurs when $n$ gets large. Indeed, dividing both sides of $n-1 \leq a_{1}n$ by $n$, we have $$1 - \frac{1}{n} \leq a_{1}.$$ Taking $n\to\infty$, this inequality becomes $1 \leq a_{1}$. This is a contradiction. – Sangchul Lee Feb 18 '14 at 18:29

Now,today I have solve this problem,I post my methods,I hope is not wrong?

since $$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$ then $$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}$$ then $$\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}$$ so $$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}$$ so $$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{1}}-\left(\dfrac{1}{a_{1}+1^2} +\dfrac{1}{a_{2}+2^2}+\cdots+\dfrac{1}{a_{n}+n^2}\right)$$ by induction,we have $$a_{n}>nt^{n+1},t=\sqrt{a_{1}}\in (0,1)$$ because $$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}>nt^{n+1}+t^{2n+2}$$ we only prove $$nt^{n+1}+t^{2n+2}>(n+1)t^{n+2},t=\sqrt{a_{1}}\in (0,1)$$ $$\Longleftrightarrow n+t^{n+1}-(n+1)t>0$$ use this Bernoulli inequality: $$(1+x)^n\ge 1+nx,x>-1,n>1$$ then we have $$t^{n+1}=(1+t-1)^{n+1}>1+(n+1)(t-1)=1+(n+1)t-(n+1)$$ Now $$\dfrac{1}{a_{n+1}}>\dfrac{1}{t^2}-\left(\dfrac{1}{t^2+1}+\dfrac{1}{2t^3+2^2}+\cdots+\dfrac{1}{nt^{n+1}+n^2}\right)$$ other hand,Use AM-GM inequality,we have \begin{align*} \dfrac{1}{t^2+1}+\dfrac{1}{2t^3+2^2}+\cdots+\dfrac{1}{nt^{n+1}+n^2}&=\dfrac{1}{1(1+t^2)}+\dfrac{1}{2(1+1+t^3)}+\cdots+\dfrac{1}{n(1+1+\cdots+1+t^{n+1})}\\ &<\dfrac{1}{2t}+\dfrac{1}{2\times 3t}+\cdots+\dfrac{1}{n(n+1)t}\\ &=\dfrac{1}{t}\left(1-\dfrac{1}{n+1}\right)\\ &<\dfrac{1}{t} \end{align*} so $$\dfrac{1}{a_{n+1}}>\dfrac{1}{t^2}-\dfrac{1}{t}$$ $$\Longrightarrow a_{n+1}<\dfrac{t^2}{1-t}$$

• +1 I checked your solution and was unable to find out any error. So, be proud of your solution! :) – Sangchul Lee Feb 19 '14 at 10:34
• The bound is actually very very good. I simulated the recurrence equation for various $a_1$ and the bound is almost touching the true steady state ,i.e. $$\lim_{n\to\infty} a_n\approx a_1\frac{1}{1-\sqrt{a_1}}$$ (see pic). Maybe there exists even an analytical solution...;) – pisoir Feb 20 '14 at 9:04