New Notes Blog: `sum_(n=0)^oo 2^n/(n!)` Use the Root Test to determine the convergence or divergence of the series.

Sunday, 30 August 2015

$\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{2}}^{{n}}}{{{n}!}}$ Use the Root Test to determine the convergence or divergence of the series.

It is usually easier to use ratio test on these types of series that contain factorials. However, we can also use root test if we rewrite factorial using exponentials. This can be accomplished using Stirling's approximation

$\displaystyle{n}!\approx\sqrt{{{2}\pi{n}}}{{\left(\frac{{n}}{{e}}\right)}}^{{n}}$

The reason why we can use this approximation is because it becomes more precise for greater values of $\displaystyle{n},$ in fact the ratio of the left and right hand side of the approximation converges to 1...

$\displaystyle{n}!\approx\sqrt{{{2}\pi{n}}}{{\left(\frac{{n}}{{e}}\right)}}^{{n}}$

$\displaystyle\lim_{{{n}\to\infty}}{\sqrt[{{n}}]{{\frac{{{2}}^{{n}}}{{n}}!}}}=\lim_{{{n}\to\infty}}{\sqrt[{{n}}]{{\frac{{{2}}^{{n}}}{{\sqrt{{{2}\pi{n}}}{{\left(\frac{{n}}{{e}}\right)}}^{{n}}}}}}}=\lim_{{{n}\to\infty}}\frac{{2}}{{{\sqrt[{{n}}]{{\sqrt{{{2}\pi{n}}}}}}\frac{{n}}{{e}}}}=$

In order to calculate $\displaystyle\lim_{{{n}\to\infty}}{\sqrt[{{n}}]{{\sqrt{{{2}\pi{n}}}}}}$ we need to use the following two facts:

$\displaystyle\lim_{{{n}\to\infty}}{\sqrt[{{n}}]{{{c}}}}={1},$ $\displaystyle{c}\in\mathbb{R}$ and $\displaystyle\lim_{{{n}\to\infty}}{\sqrt[{{n}}]{{{{n}}^{{p}}}}}={1},$ $\displaystyle{p}\in\mathbb{R}.$

Applying this to our limit yields

$\displaystyle\lim_{{{n}\to\infty}}\frac{{2}}{{{\sqrt[{{n}}]{{\sqrt{{{2}\pi{n}}}}}}\frac{{n}}{{e}}}}=\lim_{{{n}\to\infty}}\frac{{2}}{{\frac{{n}}{{e}}}}=\lim_{{{n}\to\infty}}\frac{{{2}{e}}}{{n}}=\frac{{{2}{e}}}{\infty}={0}$