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Does the series $$\sum_{n=0}^\infty\frac{\sin(n+\frac 12)\pi}{1+\sqrt{n}}$$

This is supposed to be an alternating series but I can't seem to figure out what the $b_n$ is in this case. is there some sort of sine property I am not seeing? I'm trying to test the alternating series theorem.

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  • $\begingroup$ What is $\sin \bigl((n+\frac{1}{2})\pi\bigr)$? $\endgroup$ – Daniel Fischer Jul 7 '14 at 20:11
  • $\begingroup$ im not sure what you are asking. $\endgroup$ – Joshhw Jul 7 '14 at 20:12
  • $\begingroup$ If you know what that is, everything is settled. Unless you mean the term of the series is $\frac{\pi \sin (n+\frac{1}{2})}{1+\sqrt{n}}$, in which case you will need Dirichlet's criterion. $\endgroup$ – Daniel Fischer Jul 7 '14 at 20:13
  • $\begingroup$ the term of the series is that. $\endgroup$ – Joshhw Jul 7 '14 at 20:14
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    $\begingroup$ Then I suggest moving the $\pi$ to the front, or leaving it off altogether, since it is irrelevant for convergence. If one sees $\sin (n+\frac{1}{2})\pi$, most of the time it is supposed to be parsed $\sin\bigl((n+\frac{1}{2})\pi\bigr)$. $\endgroup$ – Daniel Fischer Jul 7 '14 at 20:18
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$ \displaystyle \frac{\sin((n+\frac{1}{2})\pi)}{\sqrt{n}+1} = \frac{(-1)^n}{\sqrt{n}+1} $

So you just have to verify that $ \displaystyle \frac{1}{\sqrt{n}+1} $ decrease to zero...which should be kind of obvious.

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  • $\begingroup$ I'm trying to figure out why the top part is $(-1)^n$? is it because sin alternates between those two values and so you can treat it like that? $\endgroup$ – Joshhw Jul 7 '14 at 20:21
  • $\begingroup$ If you don't see, replace n=1,2,3,.., sin(Pi/2) = 1,sin(3Pi/2)=-1, sin(5Pi/2)=1, sin((2n+1)Pi/2) = (-1)^n ... $\endgroup$ – Dyoann Jul 7 '14 at 20:24
  • $\begingroup$ am I supposed to treat that pi as inside the parentheses? cause that would make this equation easier. $\endgroup$ – Joshhw Jul 7 '14 at 20:25
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    $\begingroup$ Please rewrite properly the problem if the pi is not inside ! But since you are talking about alternating series, it seems almost sure it's inside..otherwise it's cleary not alternate (and the Pi would be meaningless) $\endgroup$ – Dyoann Jul 7 '14 at 20:26
  • $\begingroup$ I never wrote it inside at all. that was the person responding. I'm trying to figure out if its supposed to be on the inside. $\endgroup$ – Joshhw Jul 7 '14 at 20:28
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I have the book and pi goes outside, but that is no problem.
By the alternating series test, the absolute value of
https://www.wolframalpha.com/input/?i=sin(A+%2B+B)

$\dfrac{\pi\sin(n+\dfrac{1}{2})}{1+\sqrt{n}} = \pi\dfrac{\sin(n)\cos(\dfrac{1}{2}) + \cos(n)\sin(\dfrac{1}{2})}{1+\sqrt{n}} $

is less than $\pi\dfrac{\cos(\dfrac{1}{2}) + \sin(\dfrac{1}{2})}{1+\sqrt{n}}$, which decreases and goes to zero as n approaches infinity. So, the alternating series is convergent.
In this case, we can lower the bound to $\dfrac{\pi}{1+\sqrt{n}}$ without worrying about trig identities.
We want to know whether it is conditionally or absolutely convergent.

So we let

$S_2 = \sum\dfrac{\pi|\sin(n+\dfrac{1}{2})|}{1+\sqrt{n}}$
By the ratio test, we could compare this to $ \sum\dfrac{1}{\sqrt{n}}$ and we will get a limit greater than 0.
And we know $ \sum\dfrac{1}{\sqrt{n}}$, thus $S_2$ diverges.
Finally your original series is CONDITIONALLY CONVERGENT. The best way to see this is by reading chapter 11, section 11.6 (Absolute Convergence and the Ratio and Root Test) in the book "Calculus 8th edition" by James Stewart.

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