# Ratio Test and Convergence

1. Mar 3, 2009

### 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:

$$\sum$$n=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?

2. Mar 3, 2009

### 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...

3. Mar 3, 2009

Thanks!!!