Question

Determine the convergence of the series
\[\sum_{n=1}^\infty 1-\cos(1/n)\]
(this is a problem from one of the student here)

Answer 1

is this open for anyone to answer?

Answer 2

sure

Answer 3

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.

Answer 4

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

Answer 5

two sigma notations

Answer 6

oh yep i made a mistake.

Answer 7

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

Answer 8

posted by rsvitale

Answer 9

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

Answer 10

hi watchmath can u help me after him

Answer 11

okay lol

Answer 12

watchmath can u help me

Answer 13

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

Answer 14

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

Answer 15

i mean sequence

Answer 16

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

Answer 17

so you basically guess

Answer 18

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

Answer 19

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 :).