Series; convergence, divergence

[jnimagine]

series; convergence, divergence...

Homework Statement

1. sum(infinity,n=1) n!/1.3.5...(2n-1)
2. sum(infinity, n=1) (-1)^n arcsin(-1/n)
3. sum(infinity, n=0) arcsin(1/n^2) / arctan(1/n^2)

The Attempt at a Solution

1. i used the ratio test and then i ended up with lim((n+1)(2n-1)/2n+1)) and when i solve the limit i get infinity...
but it's supposed to be convergent... so I'm obviously doing something wrong... :(
2. by alternating series test, it converges
but it's suppsoed to be conditionally convergent, so i tried using the limit test...
how do you approch this problem?? I tried using l'hospital's rule and stuff... but i ended
up with a mess lol
3. in general, when there's inverse trig functions how do u work it out?

 

[roam]

Actually for the first question [tex]\sum^{\infty}_{n=1} \frac{n!}{1.3.5...(2n-1)}[/tex] you made a mistake while simplifying:

[tex]\frac{s_{n+1}}{s_{n}} = \frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)}/\frac{n!}{1 \times 3 \times 5 \times ... \times (2n-1)}[/tex]

[tex]\frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)} \times \frac{1 \times 3 \times 5 \times ... \times (2n-1)}{n!}[/tex]


(n+1)!/n! = n+1 (which goes on top)

1x3x5x...x(2n+1) is exactly the same as 1x3x5x...x(2n-1)x(2n+1) therefore they cancel out and you are left with (2n+1) on the denomenator.

[tex]= \frac{n+1}{2n+1}[/tex]

So, [tex]lim_{n\rightarrow \infty}\frac{S_{n+1}}{S_{n}} = \frac{1}{2} < 1[/tex]

Therefore the series is...

 

[jnimagine]

Actually for the first question [tex]\sum^{\infty}_{n=1} \frac{n!}{1.3.5...(2n-1)}[/tex] you made a mistake while simplifying:

[tex]\frac{s_{n+1}}{s_{n}} = \frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)}/\frac{n!}{1 \times 3 \times 5 \times ... \times (2n-1)}[/tex]

[tex]\frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)} \times \frac{1 \times 3 \times 5 \times ... \times (2n-1)}{n!}[/tex]


(n+1)!/n! = n+1 (which goes on top)

1x3x5x...x(2n+1) is exactly the same as 1x3x5x...x(2n-1)x(2n+1) therefore they cancel out and you are left with (2n+1) on the denomenator.

[tex]= \frac{n+1}{2n+1}[/tex]

So, [tex]lim_{n\rightarrow \infty}\frac{S_{n+1}}{S_{n}} = \frac{1}{2} < 1[/tex]

Therefore the series is...

ohhhh i get it now~~ thank you! ^^

can anyone help me on inverse trig series that I've written above?? :S
also... i think i made some mistake again in using the ratio test for this question:
sum(infinity, n=1) (-1)^n((5^2n)/n!) i end up with lim(25/n+1)... and it's supposed to be convergent absolutely... so... = (

 

[Dick]

ohhhh i get it now~~ thank you! ^^

can anyone help me on inverse trig series that I've written above?? :S
also... i think i made some mistake again in using the ratio test for this question:
sum(infinity, n=1) (-1)^n((5^2n)/n!) i end up with lim(25/n+1)... and it's supposed to be convergent absolutely... so... = (

What's the limit of 25/(n+1) again? Note the parentheses, it looks pretty misleading if you leave them off. For the second one try the alternating series test. For the last one your first test for convergence should be to check that the limit of the nth term goes to zero. If it doesn't, it can't possibly converge.

 

[roam]

For the other one I think you must apply the alternating series test.

You can show using calculus (calculate a derivative) that |arcsin(-1/n)| is a decreasing function of n with limit |arcsin(0)| = 0.