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:

($$\sum$$ⁿ_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 = $$\sum$$ⁿ_n=1 $$\frac{n-(k-1)}{e^k*k^x}$$ = $$\frac{n-(k-1)}{e^n*n^x}$$

U_n+1 = $$\sum$$ⁿ_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?