Ratio Test and Convergence

[wanchosen]

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 doesnt make sense. Can I solve this for x, independent of n?

[csprof2000]

Your trouble is in the last step. You divide both sides by the log without realizing that as n->+inf, it is equal to zero.

Clearly, and you did all the work mind you, this is true for any x at all.

Reason? |x ln(n/n+1)| < 1 for all x you can think of... since any real number times zero equals zero, and this is less than 1 in absolute value.

Good job! Careful about that algebra, though...

[wanchosen]

Thanks!!!