Question

nnnn

Answer 1

\[\Large \Gamma(1/3)=\int\limits_{0}^{\infty}x^{(1/3)-1}e^{-x}dx\]

Answer 2

\[e^x=\sum_{n=1}^{\infty}\frac{x^n}{n!}\]\[e^{-x}=\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n!}\]\[x^{-2/3}e^{-x}=\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-2/3}}{n!}\]\[\Gamma(1/3)=\int\limits_{0}^{\infty}x^{-2/3}e^{-x}~dx~=~\left(\sum_{n=0}^{\infty}\frac{3(-1)^nx^{n+1/3}}{n!}+C\right){\Huge|}^{\infty}_{0}\]

Answer 3

as wolfram gives, this series sum is equivalent to
\[\Large = \lim_{x \rightarrow \infty}\left(3e^{-x}\sqrt[3]{x}\right)\]

Answer 4

and mistake, shouldn't have the +C there

Answer 5

yes, this is not a limit, this is just what it converts to.