I am trying to solve this problem from Ross's real analysis textbook, but am struggling with the later parts. Here is what I have so far.

Let $s_1 = 1$ and $s_{n+1} = \left(\frac{n}{n+1}\right)s_n^2$ for $n \geq 1$.

(a) Find $s_2$, $s_3$, and $s_4$.

(b) Show $\lim s_n$ exists.

(c) Prove $\lim s_n = 0$.


(a) This part is rather trivial, though I'll write this answer here because I reference it later. We get $s_2 = \frac{1}{2}$, $s_3 = \frac{1}{6}$, and $s_4 = \frac{1}{48}$.


(b) I believe we need to use the fact (taken as a lemma without proof, as I have proved it elsewhere) that a bounded monotonic sequence converges. By writing out terms, I know that this sequence is bounded below by $0$, as it's a product of a positive rational number and a square, and above by $1$. And, it's fairly easy to see that the sequence is decreasing, as we're scaling each term by a fraction, $\frac{n}{n+1}$. I am having difficulty, however, in proving this. One proof that I saw argued: \begin{align*} & s_{n+1} = \left(\frac{n}{n+1}\right) s_n^2 < 1 \cdot s_n^2 = s_n^2 \leq 1 \cdot s_n = s_n \\ & \therefore s_{n+1} \leq s_n, \forall n. \end{align*} But, I am having difficulty following this proof. First, this formula for $s_{n+1}$ holds only if $n > 1$. It seems that we could disregard this first term since (a) I believe we only require that the sequence decreases for an infinite number of terms and (b) we already know that $s_2 < s_2$. But, that still brings me to $s_{n+1} \leq s_n^2$, which follows from the fact that $\frac{n}{n+1} < 1$. But, I cannot understand how we get to $s_n^2 \leq s_n$: It seems that we could recursively apply our formula here, but that seems to generate an infinite loop.

Another possibility here is to write an inductive proof, but I seem to be falling into a chicken-and-egg problem as it relates to whether I ought to prove boundedness, in which case it follows that the sequence is decreasing/monotonic, or whether I should prove that it's decreasing, in which case boundedness (above by the first term) follows. It seems I would need to prove that it is bounded both below and above for the theorem to apply.


(c) Assuming we know the sequence is decreasing and bounded below (I'm fairly certain that these are the case, but struggle only with how to prove them), this sequence should converge to its infimum. It seems that this infimum is $0$, so we need to prove that $$\forall \epsilon > 0, \exists N, \forall n > N, \left \lvert s_n - 0 \right \rvert < \epsilon.$$ The only thing I can think of here is that, since $s_n$ is bounded above by $1$, $\left \lvert s_n \right \rvert \leq 1$. But this fact obviously isn't sufficient, as this must be less than any positive real number $\epsilon$. So I can't seem to figure this part out.

Any helpful insights would be greatly appreciated.


Let us look carefully at the proof given to you which you were unable to understand : $$ s_{n+1} \overset{(0)}{=} \left(\frac n{n+1}\right) s_n^2 \overset{\color{orange}{(1)}}{<} 1 \cdot s_n^2 =s_n^2 \overset{\color{blue}{(2)}}{\leq} 1 \cdot s_n = s_n $$

  • $\color{orange}{(1)}$ is true , simply because we know that the formula $(0)$ holds for $n \geq 1$, and $\frac n{n+1} = 1 - \frac 1{n+1}$, which is strictly less than $1$ whenever $n \geq 1$, and therefore $\color{orange}{(1)}$ follows by using the fact that $a < b$ and $c > 0$ implies $ac < bc$, with the appropriate $a,b,c$.

  • For $\color{blue}{(2)}$, the point is that $s_n \leq 1$ for all $n$, is already known to us. We can prove it by induction anyway : the base case $s_1 = 1$ works, and for induction, we have : $$ s_n \leq 1 \implies s_n^2 \leq 1 \implies s_{n+1} = s_n^2 \times \frac n{n+1} \leq 1 \cdot \frac n{n+1} \leq 1 $$ since $n \geq 1$ , and therefore the proof is complete.We have $s_n > 0$ from induction : the proof is very simple, I think you must have done it already. Now, again, use the fact that $a \leq b$ and $c > 0$ implies $ac \leq bc$, now with $a=c=s_n$ and $b=1$. We know that $a \leq b$ and $c > 0$, therefore $ac = s_n^2 \leq bc = 1 \cdot s_n$. This finishes the proof of $\color{orange}{(2)}$, and of the mini proof.

Boundedness both above and below were done by separate inductions : then, these bounded were used to prove that the sequence is decreasing, by using known bounds in the induction step. We are not doing anything wrong : the first two inductions help in proving the third and most important one.

For the last part, showing that $s_n$ converges to $0$ requires a re-look at the equation $(0)$, the definition of $s_n$.

The point is, the left and right hand side of this equation are two different sequences which are equated via $(0)$. Therefore, if their limits exist, they must be the same.

That is, if $\lim s_{n+1}$ exists, and if $\lim \frac n{n+1}s_n^2$ exists, then by $(0)$, these limits must be the same.

Using the usual limit criteria, one sees that (we know $s_n$ is convergent, so it has a limit) if $\lim s_n = x$, then $\lim s_{n+1} = x$, and $\lim \frac n{n+1} s_n^2 = x^2$ (using the product rule for limits, by breaking into $\frac n{n+1}, s_n$ and $s_n$, which have limits $1,x,x$ respectively).

This forces $x = x^2$, since the limits have to be the same. So $x =1$ or $x = 0$ are the only possibilities.

Once you use the decreasing nature of $s_n$ to rule out $x = 1$, your result stands before you.


Your Answer

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