A calculus problem by Rishabh Deep Singh

First we will look at the first few terms to establish a pattern:

$X_1=1\\X_2=\sqrt{1+1}=\sqrt2\\X_3=\sqrt{2+1}=\sqrt3$

By following the pattern, we see that $X_n=\sqrt{n}$ . This can be proven by mathematical induction:

Assume true for $n=k$ :

$X_k=\sqrt{k}$

Consider $n=k+1$ :

$\begin{aligned}X_{k+1}&=\sqrt{X_k^2+1}\\ &=\sqrt{\sqrt{k}^2+1}\\ &=\sqrt{k+1}\end{aligned}$

If true for $n=k$ , then true for $n=k+1$ . But $X_1=1$ , so true for $n=1$ , so by the process of mathematical induction, true for all positive integer values of $n$ .

Now we can solve the limit, as long as we know our identity for $e$ :

$\lim_{n\to\infty}{\left(\frac{X_{n+1}}{X_{n}}\right)^n}\\ =\lim_{n\to\infty}{\left(\frac{\sqrt{n+1}}{\sqrt{n}}\right)^n}\\ =\lim_{n\to\infty}{\left(\sqrt{1+\frac{1}{n}}\right)^n}\\ =\sqrt{\lim_{n\to\infty}{\left(1+\frac{1}{n}\right)^n}}\\ =\sqrt{e}\\ =1.648721270$

We note that $X_1 =1$ , $X_2 = \sqrt 2$ and $X_3 = \sqrt 3$ . It appears that we can claim that $X_n = \sqrt n$ . Let us prove the claim is true for all $n \ge 1$ by induction.

Proof: For $n=1$ , $X_1 = \sqrt 1 = 1$ as given, hence the claim is true for $n=1$ . Assuming that the claim is true for $n$ , then we have $X_{n+1} = \sqrt {X_n^2 + 1} = \sqrt {(\sqrt n)^2+1} = \sqrt{n+1}$ . Therefore the claim is also true for $n+1$ hence it is true for all $n \ge 1$ .

Then, we have:

$\begin{aligned} L & = \lim_{n \to \infty} \left(\frac {X_{n+1}}{X_n}\right)^n = \lim_{n \to \infty} \left(\frac {\sqrt{n+1}}{\sqrt n}\right)^n = \lim_{n \to \infty} \left(\sqrt{1+\frac 1n}\right)^n = \lim_{n \to \infty} \left(1+\frac 1n\right)^\frac n2 = \lim_{n \to \infty} \sqrt{\left(1+\frac 1n\right)^n} = \sqrt e \approx \boxed{1.649} \end{aligned}$