I have been stuck on this question for a while and have not got much success from class mates...can somebody please help?(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Homework Help: Ratio Test and Convergence

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