Testing for Convergence or Divergence of 1+sin(n)/10^n Series

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SUMMARY

The series from n=1 to infinity of (1 + sin(n))/10^n converges. The discussion confirms the use of the limit comparison test and the root test to establish convergence. It is established that the series can be compared to 2/10^n, which is a convergent geometric series. Therefore, the original series converges based on these tests.

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  • Understanding of series convergence tests, specifically the limit comparison test and root test.
  • Familiarity with geometric series and their convergence criteria.
  • Basic knowledge of trigonometric functions, particularly sin(n).
  • Ability to manipulate limits and series notation.
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  • Study the limit comparison test in detail, including examples and applications.
  • Learn about the root test and its conditions for series convergence.
  • Explore geometric series and their properties for convergence.
  • Investigate the behavior of trigonometric functions within series, particularly sin(n).
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Students and educators in calculus, particularly those focusing on series convergence, as well as mathematicians interested in advanced series analysis.

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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.
 
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What can you tell me of the limit of the series as n reach infinity?
 
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.
 
is the limit of the series as n goes to infinity is not 0 then the sum of the series diverge...
 
wait is it (1+sin(n))/10^n or 1+ (sin(n)/10^n)?
 
Try comparing sin n to n
 
if it's (1+sin(n))/10^n then can you tell me 1+sin(n) is smaller then what for all n?
 
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:
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
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
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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