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Testing for divergence

  • Thread starter superdave
  • Start date
  • #1
150
3

Homework Statement



Does the sum of the series from n=1 to infinity of 1+sin(n)/10^n converge or diverge.



Homework Equations





The Attempt at a Solution



I can use the comparison test or the limit comparison test.
I'm not sure where to go from here.
 

Answers and Replies

  • #2
236
0
What can you tell me of the limit of the series as n reach infinity?
 
  • #3
150
3
Well, the top part diverges, the bottom causes it to go to 0. So I don't know what happens faster.

Either it converges to 0, or it diverges.

The solution must involve the comparison test or the limit comparison test. But I'm not sure what to compare it to.
 
  • #4
236
0
is the limit of the series as n goes to infinity is not 0 then the sum of the series diverge...
 
  • #5
236
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wait is it (1+sin(n))/10^n or 1+ (sin(n)/10^n)?
 
  • #6
19
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Try comparing sin n to n
 
  • #7
236
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if it's (1+sin(n))/10^n then can you tell me 1+sin(n) is smaller then what for all n?
 
  • #8
150
3
It's (1+sin(n)). Hrm, smaller than 2. So I can compare it to 1/5^n. Now, I need to figure out how to prove that series converges. Is it a geometric series?

Actually, I know it converges, based on the root test. But I don't think we can use the root test now.
 
Last edited:
  • #9
236
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right but 1/5^n is wrong, keep it 2/10^n, now can you tell me if you know the root or the ratio test of a series?
 
  • #10
150
3
alright. So root test gives me limit of 2^1/n / 10. I don't know what 2 ^1/n goes to. Is that even possible?
 
  • #11
236
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The root test is for when n goes to infinity..
1/n~0--->2^1/n=1,so (2^1/n)/10<1
you have just now proved that the series 2/10^n converge, how can you relate this to the series you started with?
 

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