Converge or Diverge? Proving I & II Limits

[Umar]

So I have this question here from one of my assignments. I got the right answer, which is F, but I got it through eliminating other options (cos(n\pi)) diverges while sin(n\pi) obviously converges. I was thinking the only option would be IV, but the options for the answers included I and IV, or II and IV. The thing is I'm not sure how you would prove either of those to be convergent or divergent. For instance, in both of the numerators, there is (-1)^(n+1). This gives you [1,-1,1,-1...] meaning it's limit D.N.E (does not exist). The denominators are similar but different slightly. So how exactly would you go about showing that option I converges and II diverges (ie. lim as n --> infinity of those sequences)

Your help is greatly appreciated! :)

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[I like Serena]

So I have this question here from one of my assignments. I got the right answer, which is F, but I got it through eliminating other options (cos(n\pi)) diverges while sin(n\pi) obviously converges. I was thinking the only option would be IV, but the options for the answers included I and IV, or II and IV. The thing is I'm not sure how you would prove either of those to be convergent or divergent. For instance, in both of the numerators, there is (-1)^(n+1). This gives you [1,-1,1,-1...] meaning it's limit D.N.E (does not exist). The denominators are similar but different slightly. So how exactly would you go about showing that option I converges and II diverges (ie. lim as n --> infinity of those sequences)

Your help is greatly appreciated! :)

Attached is the link to the picture of the question:

Imgur: The most awesome images on the Internet

Hi Umar! ;)

Typically we try to rewrite a fraction so that it becomes clear whether it converges or not.

For (i):
$$a_n = \frac{(-1)^{n+1}n}{n^2+4} = \frac{(-1)^{n+1}}{n+\frac 4 n} \to \frac{\pm 1}{\infty + 4\cdot 0} = 0$$
So $a_n$ converges.

For (ii):
$$a_n = \frac{(-1)^{n+1}n^2}{n^2-7} = \frac{(-1)^{n+1}}{1-\frac 7 {n^2}} \to \frac{\pm 1}{1 - 7\cdot 0} = \pm 1$$
So $a_n$ diverges.

 

[Umar]

Hi Umar! ;)

Typically we try to rewrite a fraction so that it becomes clear whether it converges or not.

For (i):
$$a_n = \frac{(-1)^{n+1}n}{n^2+4} = \frac{(-1)^{n+1}}{n+\frac 4 n} \to \frac{\pm 1}{\infty + 4\cdot 0} = 0$$
So $a_n$ converges.

For (ii):
$$a_n = \frac{(-1)^{n+1}n^2}{n^2-7} = \frac{(-1)^{n+1}}{1-\frac 7 {n^2}} \to \frac{\pm 1}{1 - 7\cdot 0} = \pm 1$$
So $a_n$ diverges.

Hey! Sorry for not replying earlier but thank you so much for your help!