what method would be used to solve this infinite series?

Question

anonymous · Answer

attachment

kainui · Answer

You want to find what it converges to or if it converges?

anonymous · Answer

hehe yes @Kainui . you have been a lifesaver!

kainui · Answer

Wait lol this isn't a yes or no question I'm asking.

anonymous · Answer

haha im not quite sure !

kainui · Answer

Well find out for me please, since they are two significantly different questions and both somewhat time consuming. =)

anonymous · Answer

actually the question is asking if it converges

anonymous · Answer

and says to use any method. i assume ratio?? but not sure :|

anonymous · Answer

gosh im not thinking straight haha i reread what you were asking, idk what i was saying! its late :/

kainui · Answer

You can try ratio, but I'd probably go for the limit comparison test here or the comparison test. Why would I say that? Well it seems like I could probably just compare this to a similar function with all the constants taken away which looks like this:

$$\frac{0+\sqrt{k}}{(k+0)^3+0}=\frac{k^{1/2}}{k^3}=\frac{1}{k^{5/3}}$$

So that's sort of my strategy for thinking about comparison tests.

kainui · Answer

I could be wrong here, I just sort of made an arbitrary suggestion. I'm out of practice with these types of problems since they're usually not very important.

anonymous · Answer

hmm i thought the same method too but couldn't work through the problem :|

kainui · Answer

Wait I wrote that it became the wrong thing. It should be $$\frac{1}{k^{5/2}}$$ I don't know if that helps or not.

anonymous · Answer

oh yes i noticed that! i knew what you ment!

kainui · Answer

Yeah, I got to an answer, but it involved taking L'H's rule about 3 times haha.

kainui · Answer

$$\lim_{k \rightarrow \infty} \frac{(\frac{2+k^{1/2}}{(k+1)^3-1})}{(\frac{1}{k^{5/2}})}$$ If that's not too tiny, that's what I did.