Question about alternating series

[vande060]

Homework Statement

test the series for divergence or convergence, and the only thing we have learned so far is alternating series test

(∞, n=1) ∑ (sin(nπ)/2)/n!

Homework Equations

The Attempt at a Solution

So I know that the alternating series test has two conditions that need to be satisfied.

1. series must be decreasing
2 limit as n approaches infinity must be 0

I'm not even sure how to show this series is decreasing, I know that plugging in numbers is not enough for my prof, and I need to prove it. I am not sure how to take the limit of a factorial either; I am really lost here

Ill try to write out the first few terms just to make things more clear for myself:

Sn = 1/1 + 0 -1/6 + 0 + 1/120 ...(sin((n-1)π)/2)/(n-1)!

 

[Robert1986]

Homework Statement

test the series for divergence or convergence, and the only thing we have learned so far is alternating series test

(∞, n=1) ∑ (sin(nπ)/2)/n!

Homework Equations

The Attempt at a Solution

So I know that the alternating series test has two conditions that need to be satisfied.

1. series must be decreasing
2 limit as n approaches infinity must be 0

I'm not even sure how to show this series is decreasing, I know that plugging in numbers is not enough for my prof, and I need to prove it. I am not sure how to take the limit of a factorial either; I am really lost here

Ill try to write out the first few terms just to make things more clear for myself:

Sn = 1/1 + 0 -1/6 + 0 ...(sin((n-1)π)/2)/(n-1)!

You must mean:

$\sum{\frac{sin(\frac{n\pi}{n!})}{n!}}$

Right? Otherwise every term is 0.

So, you need to show that if this is your sequence:

$b_1 - b_2 + b_3 - b_4 \ldots$

then [itex]b_{n+1} \leq b_n$<br /> Now, forget about the n! (for now) and just write down the first few terms (actually "work out" the sin function for each term, hint: it is really easy). Now, just go and divide each term by n!. It should be clear that the sequence is decreasing. Once you have done this, try to show that limit is 0.$[/itex]

 

[vande060]

You must mean:
$\sum{\frac{sin(n\pi)}{n!}}$

yeah the alt text pi does not show up the greatest, sorry

$b_{n+1} \leq b_n$/

Im not sure how to work this out, I already wrote out the first few terms of the sequence. The numerator alternates between -1, 0, and 1. I understand that dividing by n! will show the series is decreasing, but is there a different way to show that the series is decreasing. My prof insists we prove that a series is decreasing, for example, by taking the derivative of f(x). She said that writing down the first few terms of the series is not good enough. Do you have any idea how I could do this, it doesn't even seem possible with factorials.

 

[Robert1986]

sorry, forget about my last post, I am experiencing some technical difficulties and what I wrote isn't what I meant. I'm going to try to do it again, before my connection screws up :(

 

[Robert1986]

So, what you really need to show is that 1/n! is decreasing, that is:

1/(k+1)! < 1/k!

I would re-write the denominator of 1/(k+1)! as (k+1)k!. Proving it this way is sufficient since you are doing it for any k. As for the 0's that turn up in every other term, I don't see how you can deal with this if ALL you can use is the Alternating Series Test. Because, all you can show is the the subsequence consisting of all odd n converges, unless someone else has a better idea.

 

[vande060]

So, what you really need to show is that 1/n! is decreasing, that is:

1/(k+1)! < 1/k!

I would re-write the denominator of 1/(k+1)! as (k+1)k!. Proving it this way is sufficient since you are doing it for any k.


As for the 0's that turn up in every other term, I don't see how you can deal with this if ALL you can use is the Alternating Series Test. Because, all you can show is the the subsequence consisting of all odd n converges, unless someone else has a better idea.

I understand this completely, but to prove maybe i could say

maybe i can say something like:

an = (sin(nπ)/2)/n!

bn = |an| , 0 ≤ bn ≤ 1/n!

since 1/n! is convergent, an is convergent. or is this way out there and wrong?