[tex]HOLA!![/tex]
Answer:
[tex]{\boxed{ \sum_{n=1}^{\infty}\frac{3n}{3^n \: n!}=e^{\frac{1}{3} } }}[/tex]
Explanation:
[tex]{\boxed{ \sum_{n=1}^{\infty}\frac{3n}{3^n \: n!}=}}[/tex]
For this we have to take into account:
[tex]{\boxed{ \sum_{n=1}^{\infty}\frac{x^{n} }{n!} =e^{x} }}[/tex]
Using the properties of factorials and exponents we have:
[tex]n!=(n-1)n![/tex] Also. [tex]\frac{n^{x} }{ n^{y} }=n^{x-y}[/tex]
We replace:
[tex]{\boxed{ \sum_{n=1}^{\infty}\ \frac{1}{3^{n-1}.(n-1)! } }}[/tex]
Shape it:
[tex]{\boxed{ \sum_{n=1}^{\infty}\frac{(\frac{1}{3} )^{n-1} }{(n-1)!} }}[/tex]
Finally:
[tex]{\boxed{ \sum_{n=1}^{\infty}\frac{(\frac{1}{3} )^{n-1} }{(n-1)!} =e^{\frac{1}{3} } }}[/tex]
