Convergence Criteria for Series with Exponential and Polynomial Terms

[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,

U_n = $$\frac{1}{e^n * n^x}$$

and

U_(n+1) = $$\frac{1}{e^(n+1) * (n+1)^x}$$

then

U_(n+1)/U_n = $$\frac{n^x}{e * (n+1)^x}$$

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?

 

[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!