wanchosen
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I have been stuck on this question for a while and have not got much success from class mates...can somebody please help?
Using an appropriate convergence test, find the values of x \in R for which the following series is convergent:
\sumn=1n \frac{1}{e^n * n^x}
So,
Un = \frac{1}{e^n * n^x}
and
U(n+1) = \frac{1}{e^(n+1) * (n+1)^x}
then
U(n+1)/Un = \frac{n^x}{e * (n+1)^x}
If the series is convergent:
\frac{n^x}{(n+1)^x} < e
ln(\frac{n^x}{(n+1)^x}) < 1
x * ln(\frac{n}{n+1}) < 1
n/n+1 is always positive but less than one, so the Ln term is negative,
So
x > \frac{1}{ln(n/n+1)}
But the limit of this as n tends to infininty, is infinity, which doesn't make sense. Can I solve this for x, independent of n?
Using an appropriate convergence test, find the values of x \in R for which the following series is convergent:
\sumn=1n \frac{1}{e^n * n^x}
So,
Un = \frac{1}{e^n * n^x}
and
U(n+1) = \frac{1}{e^(n+1) * (n+1)^x}
then
U(n+1)/Un = \frac{n^x}{e * (n+1)^x}
If the series is convergent:
\frac{n^x}{(n+1)^x} < e
ln(\frac{n^x}{(n+1)^x}) < 1
x * ln(\frac{n}{n+1}) < 1
n/n+1 is always positive but less than one, so the Ln term is negative,
So
x > \frac{1}{ln(n/n+1)}
But the limit of this as n tends to infininty, is infinity, which doesn't make sense. Can I solve this for x, independent of n?