## watchmath 5 years ago Determine the convergence of the series $\sum_{n=1}^\infty 1-\cos(1/n)$ (this is a problem from one of the student here)

1. anonymous

is this open for anyone to answer?

2. watchmath

sure

3. anonymous

i expanded cos(1/n) and rewrote the summand as: $\sum_{n=1}^{\infty}(-1)^{n+1}/(2n)!n^{2n}$ which is an alternating series that is monotone decreasing, so it converges.

4. watchmath

you mean using the Taylor series? then shouldn't we have to sigma notation?

5. watchmath

two sigma notations

6. anonymous

oh yep i made a mistake.

7. anonymous

hey ty for posting it up the question but whats the answer is ∑n=1∞(−1)n+1/(2n)!n2n

8. anonymous

posted by rsvitale

9. watchmath

no, he made a mistake. I'll post the solution if nobody answer this :).

10. toxicsugar22

hi watchmath can u help me after him

11. anonymous

okay lol

12. toxicsugar22

watchmath can u help me

13. anonymous

by the way do you know how to find the general term of a series

14. anonymous

if given something like this 3,6,10,15

15. anonymous

i mean sequence

16. watchmath

you need to try to see the pattern and formulate it in general.

17. anonymous

so you basically guess

18. watchmath

No guessing is not the right word for it. It is observing some regularity :).

19. watchmath

Ok, let solve the series problem. By the double angle formula we know that 1- cos(2x)= 2sin^2(x) It follows that $$1-\cos(1/n)=2\sin^2(1/(2n))$$ Recall that for positive x we have $$\sin x < x$$ It follows that $$\sin^2(1/(2n))< (1/(2n) )^2$$ But know the series $$\sum 1/(2n)^2$$ is convergent (p-series with p=2). Hence $$\sum 2\sin^2(1/(2n))$$ is convergent as well :).