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Convergence: Root Test Inconclusive

  • Thread starter JRangel42
  • Start date
  • #1
17
0

Homework Statement



I'm this series to see if it's convergent or divergent. I tried the root test, but it came out inclusive, and now I am trying to figure out if the ration test works. The only thing I'm asking which would be the right test for this.

Homework Equations



∞Ʃn=1 (2n+1)^n/(n^2n)

The Attempt at a Solution



[(2n+1)^n/(n^2)]^1/n
= (2n+1)/(n^2)
= [itex]\stackrel{}{lim n\rightarrow}[/itex]∞ (2n+1)/(n^2)
= 1/1
= 1
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Oh, come on. The limit of (2n+1)/n^2 is not 1.
 
  • #3
17
0
Oh, jeez! You're right! I made a dumb mistake and forgot to divide the 1 in (2n+1) by n^2. Oh, and thanks.
 
  • #4
hello, i have another problem. Could you help me ?
The problem is ;

∞Ʃn=0 (1+1/n)^n

When I use root theorem;

n√((1+1/n)^n) = (1+1/n)

lim n→∞ (1+1/n) = 1

Result is inconclusive so what should I do ?

thanks in advance
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
618
hello, i have another problem. Could you help me ?
The problem is ;

∞Ʃn=0 (1+1/n)^n

When I use root theorem;

n√((1+1/n)^n) = (1+1/n)

lim n→∞ (1+1/n) = 1

Result is inconclusive so what should I do ?

thanks in advance
Try the nth term test. http://en.wikipedia.org/wiki/Term_test
 
  • #6
Ok I tried it:

lim n→∞ (1+(1/n))^n = e (it is written in some theorem)

And rule of nth term test is
If lim n→∞ an ≠0 or if the limit does not exist, then ∞Ʃn=1 an diverges.

My result is 'e' is not equal 0 so, ∞Ʃn=1 an diverges.

My original question :

∞Ʃn=0 an
Is there any problem that in my question the sigma starts with 0 but nth term test rule says sigma starts with 1.

Does it change anything ?

If not, my result is divergence.
 
  • #7
Dick
Science Advisor
Homework Helper
26,258
618
Ok I tried it:

lim n→∞ (1+(1/n))^n = e (it is written in some theorem)

And rule of nth term test is
If lim n→∞ an ≠0 or if the limit does not exist, then ∞Ʃn=1 an diverges.

My result is 'e' is not equal 0 so, ∞Ʃn=1 an diverges.

My original question :

∞Ʃn=0 an
Is there any problem that in my question the sigma starts with 0 but nth term test rule says sigma starts with 1.

Does it change anything ?

If not, my result is divergence.
No, it doesn't change anything, and that's an important point. All that matters for convergence or divergence is what happens 'near infinity'. Changing any finite number of terms of your series doesn't affect convergence.
 
  • #8
Ok, I got it. Thank you so much.
 

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