Using an appropriate convergence test, find the values of x [tex]\in[/tex]

**for which the following series is convergent:**

*R*[tex]\sum[/tex]

_{n=1}

^{n}[tex]\frac{1}{e^n * n^x}[/tex]

So,

U

_{n}= [tex]\frac{1}{e^n * n^x}[/tex]

and

U

_{(n+1)}= [tex]\frac{1}{e^(n+1) * (n+1)^x}[/tex]

then

U

_{(n+1)}/U

_{n}= [tex]\frac{n^x}{e * (n+1)^x}[/tex]

If the series is convergent:

[tex]\frac{n^x}{(n+1)^x}[/tex] < e

ln([tex]\frac{n^x}{(n+1)^x}[/tex]) < 1

x * ln([tex]\frac{n}{n+1}[/tex]) < 1

n/n+1 is always positive but less than one, so the Ln term is negative,

So

x > [tex]\frac{1}{ln(n/n+1)}[/tex]

But the limit of this as n tends to infininty, is infinity, which doesnt make sense. Can I solve this for x, independent of n?