Could you prove convergence?

Question

Two questions:

Prove $a_n$ is bounded by 2 if $a_1=0, a_2=\sqrt2,\ldots,a_{n+1}=\sqrt{2+a_n}$

Prove (I've already proven $s_n$ is monotonically increasing and bounded above by 3) $\lim_{n \to \infty} s_n=e$ for $s_n=(1+\frac1n)^n$.

Special case of $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ — YuiTo Cheng, Jul 11 '19 at 08:00

score 10 · Answer 1 · edited Apr 13 '17 at 12:19

10

It is true for $a_1=0$. Now suppose true for $a_n$. Then $a_{n+1}=\sqrt{2+a_n}\leq \sqrt{2+2}=2$ so by induction this is true for every $n\geq 1$.

How are you defining $e$ for the second question? Sometimes that is just the definition! See this for example, or this for such "definition depending" problems.

edited Apr 13 '17 at 12:19

Community

1

answered Jun 23 '13 at 02:43

Pedro

122,002

score 1 · Answer 2 · answered Jun 23 '13 at 06:04

As Peter already said in his answer, your second question's answer may depend on how you define the number $\;e\;$, yet you can also check the following function's limit:

$$\lim_{x\to\infty}\log\left(1+\frac1x\right)^x=\lim_{x\to\infty}x\log\left(1+\frac1x\right)=\lim_{x\to\infty}\frac{\log\left(1+\frac1x\right)}{\frac1x}\stackrel{\text{l'Hospital}}=$$

$$=\lim_{x\to\infty}\frac x{x+1}=1$$

and from here you get what you want.

Could you prove convergence?

2 Answers2