Series Convergence Test

  • Thread starter wanchosen
  • Start date
  • #1
19
0
I am having problems with the following question:

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

([tex]\sum[/tex]nk=1 1/ekkx)n

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

n1 = [tex]\frac{1}{e}[/tex]

n2 = [tex]\frac{1}{e}[/tex] + [tex]\frac{1}{e^2 * 2^x}[/tex]

n3 = [tex]\frac{1}{e}[/tex] + [tex]\frac{1}{e^2 * 2^x}[/tex] + [tex]\frac{1}{e^3 * 3^x}[/tex]

n4 = [tex]\frac{1}{e}[/tex] + [tex]\frac{1}{e^2 * 2^x}[/tex] + [tex]\frac{1}{e^3 * 3^x}[/tex] + [tex]\frac{1}{e^4 * 4^x}[/tex]

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

So,

Un = [tex]\sum[/tex]nn=1 [tex]\frac{n-(k-1)}{e^k*k^x}[/tex] = [tex]\frac{n-(k-1)}{e^n*n^x}[/tex]

Un+1 = [tex]\sum[/tex]nk=1 [tex]\frac{(n+1)-(k-1)}{e^n*n^x}[/tex] = [tex]\frac{n-k+2}{e^(n+1) * (n+1)^x}[/tex]

n-->[tex]\infty[/tex]

Un+1/Un = [tex]\frac{(n-k+2)n^x}{(n-k+1)*(n+1)^x}[/tex]

divide by n

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

Lim[tex]\infty[/tex] nx-1 if convergent

|nx-1| < 1

so,

x-1 < 0

x < 1

Does this look right?
 

Answers and Replies

  • #2
35,235
7,054
How did you get this?
[tex]\sum all n = \frac{n}{e} + \frac{n - 1}{e^2 * 2^x} + \frac{n - 2}{e^3 * 3^x} [/tex]
 
  • #3
19
0
I assumed because the whole expression is encased in ()n that we summed all terms for each value of k upto n.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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
[tex]\left(e^{-k}k^{-x}\right)\left(e^{k+1}(k+1)^{x}\right)= e\left(\frac{k+1}{k}\right)^k[/tex]
What is that as k goes to infinity? For what x is it less than 1?
 
  • #5
19
0
Yes, I had misunderstood the ()n.

I believe, effectively the series is :-

[tex]\sum[/tex]n [tex]\frac{1}{e^n * n^x}[/tex]

If

Un = [tex]\frac{1}{e^n * n^x}[/tex]

and

U(n+1) = [tex]\frac{1}{e^(n+1) * (n+1)^x}[/tex]

then

U(n+1)/Un = [tex]\frac{n^x}{e * (n+1)^x}[/tex]

Therefore if the series is convergent:

[tex]\frac{n^x}{(n+1)^x}[/tex] < e

ln([tex]\frac{n^x}{(n+1)^x}[/tex]) < 1

x * ln([tex]\frac{n}{n+1}[/tex]) < 1

n/n+1 is always positive but less than one :- ln term is negative,

x > [tex]\frac{1}{ln(n) - ln (n+1)}[/tex]

but is this a final solution?
 

Related Threads on Series Convergence Test

  • Last Post
2
Replies
28
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
2K
Replies
17
Views
918
Replies
1
Views
882
Replies
4
Views
739
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
898
  • Last Post
Replies
2
Views
1K
Top