1. ## series

$\sum_{k\ =\ 0}^n (cos 1)^k$

i'm supposed to tell if this converges or diverges, & if it converges, evaluate the sum

i know that cosine oscillates..but because it's cosine of a specific value, i'm a little confused. since cos 1 is about 1, will this just go up to infinity? and if so, how would i show it?

2. Hi

$\sum_{k\ =\ 0}^n q^k = \frac{1-q^{n+1}}{1-q}$

When |q| < 1 the series converge to $\frac{1}{1-q}$

This is the case since $|cos 1| < 1$

3. Originally Posted by buttonbear
$\sum_{k\ =\ 0}^n (cos 1)^k$

i'm supposed to tell if this converges or diverges, & if it converges, evaluate the sum

i know that cosine oscillates..but because it's cosine of a specific value, i'm a little confused. since cos 1 is about 1, will this just go up to infinity? and if so, how would i show it?
At first glance I would replace cos1 with its Taylor series and you could make a few quick observations as to whether the series converges or diverges.

4. Running-Gag's method is a lot better.

5. i haven't actually learned taylor series yet..

is the first method using the definition of geometric series?

6. Originally Posted by buttonbear
i haven't actually learned taylor series yet..

is the first method using the definition of geometric series?
Did you really read running-gag's post ?

$\sum_{k= 0}^n q^k = \frac{1-q^{n+1}}{1-q}$
When $|q| < 1$ the series converges to $\frac{1}{1-q}$

This is the case since $|cos 1| < 1$
what is this ?

7. there's really no reason to be rude. it looks similar to the definition for the geometric series, where the sum converges to a/1-r, but not identical. i haven't seen the other form that running-gag posted, so i was just asking. i'm here to understand, if you don't want to help me, you don't have to.

8. Originally Posted by buttonbear
there's really no reason to be rude. it looks similar to the definition for the geometric series, where the sum converges to a/1-r, but not identical. i haven't seen the other form that running-gag posted, so i was just asking. i'm here to understand, if you don't want to help me, you don't have to.
Did I look rude ?? It wasn't my intention. And I tried to help you, as I didn't give immediately the solution, but made you think about it...

And what is a in your formula ? It's the first term of the series ! (it should be better with a smilie ><)
Your series starts at k=0, and when k=0, $[\cos(1)]^k=1$
So a=1.

9. i'm sorry if you weren't trying to be, it just seemed like it because you were like "did you actually read it??" >.< i've had a couple of bad experiences on here, so, i just wanted to make sure. i appreciate your help, and, i agree that it's better when you don't give the answer right away. thanks again, sorry for the misunderstanding.

edit:
the series actually started at k=1, so, that's probably where i'm getting hung up..

10. Originally Posted by buttonbear
edit:
the series actually started at k=1, so, that's probably where i'm getting hung up..
Do you mean that the series is actually $\sum_{k\ =\ 1}^n (cos1)^k$ ?

$\sum_{k\ =\ 1}^n q^k = \left(\sum_{k\ =\ 0}^n q^k\right) - 1 = \frac{1-q^{n+1}}{1-q} - 1 = \frac{q-q^{n+1}}{1-q}$

11. yes, that's what i mean

it's actually $\sum_{k\ =\ 1}^{\infty} (cos1)^k$, though i don't know how much that infinity affects it

so, you're saying you can just subtract 1 from the rest of the series to reconcile the difference in starting value of k?

12. Originally Posted by buttonbear
yes, that's what i mean

it's actually $\sum_{k\ =\ 1}^{\infty} (cos1)^k$, though i don't know how much that infinity affects it

so, you're saying you can just subtract 1 from the rest of the series to reconcile the difference in starting value of k?
Or you can factorise :
$\sum_{k=1}^\infty (\cos(1))^k=\cos(1) \sum_{k=1}^\infty (\cos(1))^{k-1}$
if you let n=k-1 (for more convenience), you get :
$\cos(1) \sum_{n=0}^\infty (\cos(1))^n$

Anyway, you can just apply the formula $\frac{a}{1-q}$
If the series starts at k=1, then the first term $a=\cos(1)^1=\cos(1)$
that's all

13. i'm sorry that i'm having so much trouble getting my head around this, but, what is q, in this case? i guess i'm confused by the whole n=k-1 thing.. i'm not sure how it directly converts to the formula a/1-r

14. a is the first term and q is the ratio (2nd term is aq, third term is aq² etc)

$\sum_{k\ =\ 0}^{\infty} aq^k = a\:\sum_{k\ =\ 0}^{\infty} q^k = \frac{a}{1-q}$

When taking the sum from 1 and not from 0 you can
either take away the first term (a)
$\sum_{k\ =\ 1}^{\infty} aq^k = \left(\sum_{k\ =\ 0}^{\infty} aq^k\right) - a = \frac{a}{1-q} - a = \frac{aq}{1-q}$

or factor q
$\sum_{k\ =\ 1}^{\infty} aq^k = q \sum_{k\ =\ 1}^{\infty} aq^{k-1} = q \sum_{j\ =\ 0}^{\infty} aq^j = q\:\frac{a}{1-q} = \frac{aq}{1-q}$

Fortunately the result is the same

15. that's a really helpful explanation, thank you!

so, for this problem, i ended up with $\frac{cos1}{1-cos1}$ for the sum, because i said a= cos1 and then to get to the next term you multiply it by itself, so r=cos1 also?

is this correct?

Page 1 of 2 12 Last