Question

Is the following series divergent or convergent?

Answer 1

\[\sum_{p=1}^{\infty} \frac{ psin(p+1)\frac{ \Pi }{ 2} }{ (p+100) }\]

Answer 2

@rational

Answer 3

Try divergence test

Answer 4

That should do it I guess

Answer 5

I'm trying the ratio test but it is really complicated and very hard to simplify

Answer 6

Yeah not sure how that would work out, but it would get messy quick try divergent test we have \[\lim_{p \rightarrow \infty} a_p \neq 0 \implies \sum a_p ~~ diverges\]

Answer 7

How would I do this? sorry I have never seen this topic before

Answer 8

Let \[a_p = \frac{ p \sin(p+1)\frac{ \pi }{ 2 } }{ (p+100) }\] then take the limit \[\lim_{p \rightarrow \infty} a_n = \lim_{p \rightarrow ?}\]

Answer 9

\[\lim_{p \rightarrow \infty} a_p = \lim_{p \rightarrow \infty} \frac{ p \sin(p+1)\frac{ \pi }{ 2 } }{ (p+100) }\]

Answer 10

You get pi/2 so it diverges

Answer 11

Why pi/2 ? do all of the p values go to infinity, therefore you're left with pi/2?
I thought the limit has an actual value it converges?

Answer 12

I think it is converge to 0
Let's do some terms:
if p =1, the first term of the sum is 0 since sin (1+1) pi/2 = sin pi =0
if p =2, then the second term is -1/51
the third term is 0 (sin 2pi =0)
the fourth term is 1/105
the fifth term is 0
the sixth one is -1/107
the seventh one is 0
the eighth one is 1/109
Hence for p is odd, the term = 0
But for p is even, the term goes from positive to negative and monotonous decreases to 0 but never = 0.
Hence the sum converges to 0

Answer 13

okay but the answer somehow says that it is divergent?

Answer 14

@Loser66 I'm thinking perhaps that for it to converge, the rule applies that a1>a2>a3>a4, and in this case every other term is zero so it does not follow the rule??? Would this sound correct?

Answer 15

The sum will reduce to another sum like \(\sum_{p=1}^{\infty} \dfrac{2p sin(2p+1) \pi/2}{2p+100}\) to get rid off all of 0 terms. Then, it works well