Convergence Test for Series with Exponential and Polynomial Terms

[wanchosen]

I am having problems with the following question:

Using an appropriate convergence test, find the values of x \in R for which the following series is convergent:

(\sumⁿ_k=1 1/e^kk^x)_n

I used the ratio test to solve this but I'm not so sure about my solution:

n₁ = \frac{1}{e}

n₂ = \frac{1}{e} + \frac{1}{e^2 * 2^x}

n₃ = \frac{1}{e} + \frac{1}{e^2 * 2^x} + \frac{1}{e^3 * 3^x}

n₄ = \frac{1}{e} + \frac{1}{e^2 * 2^x} + \frac{1}{e^3 * 3^x} + \frac{1}{e^4 * 4^x}

\sum all n = \frac{n}{e} + \frac{n-1}{e^2*2^x} + \frac{n-2}{e^3*3^x}

So,

U_n = \sumⁿ_n=1 \frac{n-(k-1)}{e^k*k^x} = \frac{n-(k-1)}{e^n*n^x}

U_n+1 = \sumⁿ_k=1 \frac{(n+1)-(k-1)}{e^n*n^x} = \frac{n-k+2}{e^(n+1) * (n+1)^x}

n-->\infty

U_n+1/U_n = \frac{(n-k+2)n^x}{(n-k+1)*(n+1)^x}

divide by n

= \frac{((1-(k/n))n^(x-1)}{(1-(k/n)+(1/n)*((n+1)^x)/n}

Lim_\infty n^x-1 if convergent

|n^x-1| < 1

so,

x-1 < 0

x < 1

Does this look right?

 

[Mark44]

How did you get this?
\sum all n = \frac{n}{e} + \frac{n - 1}{e^2 * 2^x} + \frac{n - 2}{e^3 * 3^x}

 

[wanchosen]

I assumed because the whole expression is encased in ()_n that we summed all terms for each value of k upto n.

 

[HallsofIvy]

I'm not sure why you are calling the "n₁", "n₂", etc. I presume you just mean "n= 1", "n= 2", etc.

And, I'm not at all clear on what you are doing calculating all those terms. You said you used the ratio test but that's just
\left(e^{-k}k^{-x}\right)\left(e^{k+1}(k+1)^{x}\right)= e\left(\frac{k+1}{k}\right)^k
What is that as k goes to infinity? For what x is it less than 1?

 

[wanchosen]

Yes, I had misunderstood the ()_n.

I believe, effectively the series is :-

\sum_n \frac{1}{e^n * n^x}

If

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}

Therefore 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 :- ln term is negative,

x > \frac{1}{ln(n) - ln (n+1)}

but is this a final solution?