Question

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Answer 1

$HOLA!!$

Answer:

${\boxed{ \sum_{n=1}^{\infty}\frac{3n}{3^n \: n!}=e^{\frac{1}{3} } }}$

Explanation:

${\boxed{ \sum_{n=1}^{\infty}\frac{3n}{3^n \: n!}=}}$

For this we have to take into account:

${\boxed{ \sum_{n=1}^{\infty}\frac{x^{n} }{n!} =e^{x} }}$

Using the properties of factorials and exponents we have:

          $n!=(n-1)n!$  Also.    $\frac{n^{x} }{ n^{y} }=n^{x-y}$  

We replace:

${\boxed{ \sum_{n=1}^{\infty}\ \frac{1}{3^{n-1}.(n-1)! } }}$

Shape it:

${\boxed{ \sum_{n=1}^{\infty}\frac{(\frac{1}{3} )^{n-1} }{(n-1)!} }}$

Finally:

${\boxed{ \sum_{n=1}^{\infty}\frac{(\frac{1}{3} )^{n-1} }{(n-1)!} =e^{\frac{1}{3} } }}$