Question

How to use test for divergence? `sum_(n=2) ^oo ((4+n)/n)^n`

`sum_(n=2) ^oo ((4+n)/n)^n`

Answer 1

The series diverges.

Explanation:

I would use the limit test for divergence.

`L = lim_(n -> oo) ((4 + n)/n)^n`

`L = lim_(n-> oo) (4/n + 1)^n`

`lnL = lim_(n-> oo) ln(4/n + 1)^n`

`lnL =lim_(n -> oo) n ln(4/n +1)`

`lnL = lim_(n -> oo) ln(4/n + 1)/(1/n)`

Since we get `0/0` evaluting these limits, we can use l'hospitals rule.

`lnL = lim_(n-> oo) ((-4/n^2)/(4/n + 1))/(-1/n^2)`

Tidying up a little, we get:

`lnL = lim_(n-> oo) (4n)/(n + 4)`

`lnL =lim_(n -> oo) ((4n)/n)/(n/n + 4/n)`

`lnL = 4/1`

`L = e^4`

Since `e^4 != 0`, this series diverges.

Hopefully this helps!