$\displaystyle lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n*(\frac{2n+1}{n+1})^n$
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$\displaystyle lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n*(\frac{2n+1}{n+1})^n$
Note that:
$\displaystyle \left(\frac{n+1}{2n}\right) \left(\frac{2n+1}{n+1}\right)$
$\displaystyle =\frac{2n+1}{2n}=1+\frac{1}{2n}$
$\displaystyle \lim_{n\to \infty}\left(\frac{2n+1}{2n}\right)^{n}=\lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$
Notice the similarity to the famous 'e' limit?.
$\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{n}=e$
So, what is the limit of $\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$
Hint :
$\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?$
Quote:
Originally Posted by Raoh
$\displaystyle e^{\frac{1}{2}}$
but, what about :
$\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n$
Quote:
Originally Posted by Raoh
$\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}$
therefore your limit must be $\displaystyle e^{\frac{1}{2}}$.Quote:
Originally Posted by Raoh
(Happy)
O.k but what about
$\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$
i believe (Happy)Quote:
Originally Posted by gilyos
$\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$
$\displaystyle
\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n = 0
$
Why?
Intuition:
$\displaystyle
(\frac{1}{2} + \frac{1}{2n})
$
is always less than 1, and then raise it to any power greater than one, and it will get smaller.
Given the discussion in this thread, a bit of thought might have suggested writing $\displaystyle \left( \frac{1}{2} + \frac{1}{2n}\right)^n = \left( \frac{1}{2}\right)^n \left( 1 + \frac{1}{n}\right)^n$ from which the limit is obvious.Quote:
Originally Posted by gilyos
