Let $\displaystyle s_{1}=\sqrt{p}, $where$\displaystyle p>0 $and
$\displaystyle s_{n+1}=\sqrt{p+s_n}
$for n in naturals
Find the limit of $\displaystyle s_{n}$ Hint: One upper bound of the sequence is $\displaystyle 1+2\sqrt{p}$
Let $\displaystyle s_{1}=\sqrt{p}, $where$\displaystyle p>0 $and
$\displaystyle s_{n+1}=\sqrt{p+s_n}
$for n in naturals
Find the limit of $\displaystyle s_{n}$ Hint: One upper bound of the sequence is $\displaystyle 1+2\sqrt{p}$
The ACTUAL fact us that $\displaystyle \sum^\infty_{n=0}\frac{1}{n!}=e$ ...the sum is infinite, and either one defines the number e that way, or else defines some other way and proves the above sum is e. Anyway, any decent calculus book must include a discussion on this. For example, Apostol's "Calculus I" has several interesting parts on the number e, including its definition as that number x for which $\displaystyle \ln x = 1$ .
Tonio
Let $\displaystyle L $ be the limit of the sequence $\displaystyle s_n $.
Since $\displaystyle s_{n}^2=p+s_{n-1} $ and for large enough $\displaystyle N\; |s_n-s_{n-1}|<\epsilon $, $\displaystyle L^2=p+L $.
Hence $\displaystyle L $ is the positive solution to $\displaystyle x^2-x-p=0 $.
Thus $\displaystyle L = \frac{1+\sqrt{4p+1}}{2} $.