Convergence of 2 + and 1 - alternating series

Click For Summary
SUMMARY

The forum discussion centers on the convergence of the alternating series defined as 1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + ... Participants explored the application of the Leibniz rule and ultimately determined that the series diverges. The transformation of the series into f(n) = 1/n - 1/(n+1) + 1/(n+2) was analyzed, leading to a comparison test that confirmed divergence due to the behavior of f(n) as n approaches infinity. The conclusion drawn is that the series diverges based on the comparison test with f(n) being greater than C/n for some constant C.

PREREQUISITES
  • Understanding of alternating series and convergence tests
  • Familiarity with the Leibniz rule for alternating series
  • Knowledge of comparison tests in series convergence
  • Basic calculus concepts, including limits and asymptotic behavior
NEXT STEPS
  • Study the application of the Leibniz rule for alternating series
  • Learn about the comparison test for series convergence
  • Explore the behavior of functions as n approaches infinity
  • Investigate other convergence tests such as the ratio test and root test
USEFUL FOR

Students studying calculus, particularly those focusing on series and convergence, as well as educators seeking to clarify concepts related to alternating series and convergence tests.

benf.stokes
Messages
67
Reaction score
0

Homework Statement



The problem asks you if the series:
1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}...

converges or diverges

Homework Equations


The Attempt at a Solution



I tried to apply the Leibniz rule but I realized it can't be applied. Is there a transformation of this series that enables the use of Leibniz rule?
 
Physics news on Phys.org
If you write f(n)=1/n-1/(n+1)+1/(n+2), then the series becomes 1+f(2)+f(5)+f(8)+... Can you say something about the behavior of f(n) that might make you think it converges or diverges?
 
lim n tend to infinity of f(n) is different from zero and thus the series does not converge?
 
benf.stokes said:
lim n tend to infinity of f(n) is different from zero and thus the series does not converge?

Why do you think lim n->infinity f(n) isn't zero??
 
Sorry. I misread your post. It is indeed zero. How then is this rewriting useful? f(n) is always positive and the ratio test is useless as it gives r=1
 
benf.stokes said:
Sorry. I misread your post. It is indeed zero. How then is this rewriting useful? f(n) is always positive and the ratio test is useless as it gives r=1

Try a comparison test. Can you make a simple estimate of the size of f(n) for large n? Combine the terms in f(n).
 
f(n)=\frac{n^2+2*n+6}{n*(n+1)*(n+2)}=\frac{n+1}{n*(n+2)}+\frac{5}{n*(n+1)*(n+2)} > \frac{n+1}{n*(n+2)}>\frac{1}{n}

and thus by the comparison test the series diverges?
 
benf.stokes said:
f(n)=\frac{n^2+2*n+6}{n*(n+1)*(n+2)}=\frac{n+1}{n*(n+2)}+\frac{5}{n*(n+1)*(n+2)} > \frac{n+1}{n*(n+2)}>\frac{1}{n}

and thus by the comparison test the series diverges?

I think your comparisons might be a bit sloppy. But you've got an n^2 in the numerator and an n^3 in the denominator, so you can certainly show f(n)>C/n for some value of C. You also have a be a little careful because the series doesn't sum over all n. Only over n=2,5,8,etc. So you'll need an argument of show that series diverges as well. But sure, once you have f(n)>C/n, the series diverges.
 
Thanks for everything . One last thing though: how should I do the comparison so that it's less sloppy?
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K