Question

`\huge{\boxed{ \sum_(n=1)^(\infty)(3n)/(3^n \: n!)=}}`

Answer 1

`HOLA!!`

Answer:

`{\boxed{ \sum_(n=1)^(\infty)(3n)/(3^n \: n!)=e^{(1)/(3) } }}`

Explanation:

`{\boxed{ \sum_(n=1)^(\infty)(3n)/(3^n \: n!)=}}`

For this we have to take into account:

`{\boxed{ \sum_(n=1)^(\infty)(x^(n) )/(n!) =e^(x) }}`

Using the properties of factorials and exponents we have:

`n!=(n-1)n!` Also.
`(n^(x) )/( n^(y) )=n^(x-y)`

We replace:

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

Shape it:

`{\boxed{ \sum_(n=1)^(\infty)(((1)/(3) )^(n-1) )/((n-1)!) }}`

Finally:

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