Test for convergence

[Guest]

I need to determine whether it converges absolutely, conditionally or diverges.

I'm stuck on this problem, not sure how to start it, do I split up the fraction?

Sigma as n=1 to infinity: (cos(n*pi / 2) * sin( n*pi - pi/2)) / (sqrt(n) + 1)

 

[skeeter]

evaluate and list out the first few terms of the series (about 8 of them) ... see what you get.

 

[Guest]

evaluate and list out the first few terms of the series (about 8 of them) ... see what you get.

Ok, I'm getting that it is decreasing and that the limit is 0. Is this right?

So I took the absolute value of it, diverges.

So it converges conditionally.

Is this right?

 

[soroban]

Hello, 27077!

Ok, I'm getting that it is decreasing and that the limit is 0. Is this right? . . . Right!

So I took the absolute value of it, it diverges . . . Correct!

So it converges conditionally . . . Yes!

skeeter's approach is direct and much easier.

I used some Trig . . . with the same results.


Since \(\displaystyle \,\sin(A\,-\,B)\:=\:\sin(A)\cos(B)\,-\,\sin(B)\cos(A)\)

\(\displaystyle \;\;\)then: \(\displaystyle \,\sin\left(n\pi\,-\,\frac{\pi}{2}\right)\:=\:\sin(n\pi)\cdot\cos\left(\frac{\pi}{2}\right)\,-\,\sin\left(\frac{\pi}{2}\right)\cdot\cos(n\pi) \;= \;-\cos(n\pi)\)


Hence: \(\displaystyle \L\,\sum^{\infty}_{n=1}\,\frac{\cos\left(n\cdot\frac{\pi}{2}\right)\cdot\sin\left(n\pi\,-\,\frac{\pi}{2}\right)}{\sqrt{n}\,+\,1} \;=\;\sum^{\infty}_{n=1}\,-\frac{\cos\left(n\cdot\frac{\pi}{2}\right)\cdot\cos(n\pi)}{\sqrt{n}\,+\,1}\)


We find that:

\(\displaystyle \;\;\)For odd \(\displaystyle n:\;\cos\left(n\cdot\frac{\pi}{2}\right)\,=\,0\)

\(\displaystyle \;\;\)For even \(\displaystyle n:\:\cos\left(n\cdot\frac{\pi}{2}\right) \,=\,\pm1\) and \(\displaystyle \cos(2n\pi)\,=\,1\)


Therefore, the series is: \(\displaystyle \L\,\frac{1}{\sqrt{2}+1}\,-\,\frac{1}{\sqrt{4}+1} \,+\,\frac{1}{\sqrt{6}+1} \,-\,\frac{1}{\sqrt{8}+1} \,+\,\cdots\)

\(\displaystyle \;\;\\)which is indeed conditionally convergent.