Convergence of sum using D'Alembert.

Question

I have to find the convergence of this series: $$\sum \limits_{n=0}^{\infty} \frac{(1+{\frac 1n})^n}{2^n}$$ I started by using D'Alembert: $\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n}$, So : $$\lim \limits_{n \to \infty} \frac {\frac{(1+{\frac {1}{n+1}})^{n+1}}{2^{n+1}}}{\frac{(1+{\frac {1}{n}})^{n}}{2^{n}}}=$$ $$\lim \limits_{n \to \infty} \frac{(1+{\frac {1}{n+1}})^{n+1}}{2^{n+1}}*\frac{2^{n}}{(1+{\frac {1}{n}})^{n}}=$$ I am stuck at this point because I don't really know how to do my simplifications. Any help would be great! Thanks.

Answer 1

$$ \lim \limits_{n \to \infty} \frac{(1+{\frac {1}{n+1}})^{n+1}}{2^{n+1}}*\frac{2^{n}}{(1+{\frac {1}{n}})^{n}}= \lim \limits_{n\to\infty}\frac{(1+{\frac{1}{n+1}})^{n+1}}{2^{1}\times(1+{\frac {1}{n}})^{n}} $$ Since $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e$ (the $+1$ in the numerator doesn't make a difference, as we are dealing with infinity): $$ =\lim \limits_{n\to\infty}\frac{e}{2e}=\frac{1}{2} $$

Answer 2

Hint: For $n>1$ you have: $$0\le\frac{(1+{\frac 1n})^n}{2^n}= \left(\frac{1}{2}+\frac{1}{2n}\right)^n \le \left(\frac{3}{4}\right)^n $$