$$\sum_{n=1}^{\infty} \frac{n^{2}+1}{n^{3} +1}$$
the way I would solve it is
$$\frac{n^{2}/n^{2}+1/n^{2}}{n^{3}/n^{2} +1/n^{2}}$$
$$\frac{1+1/n^{2}}{n+1/n^{2}}$$
and as $$ n\rightarrow \infty$$ we get $$1/\infty $$ =0
put the answer is: divergent, so I guess my calculations are wrong

Question

$$\sum_{n=1}^{\infty} \frac{n^{2}+1}{n^{3} +1}$$
the way I would solve it is
$$\frac{n^{2}/n^{2}+1/n^{2}}{n^{3}/n^{2} +1/n^{2}}$$
$$\frac{1+1/n^{2}}{n+1/n^{2}}$$
and as $$ n\rightarrow \infty$$ we get $$1/\infty $$ =0
put the answer is: divergent, so I guess my calculations are wrong

anonymous · Answer

Can you take n^3 common from both numerator and denominator???

anonymous · Answer

you mean in stead of n^3?

anonymous · Answer

i mean n^2

anonymous · Answer

Yes please...

anonymous · Answer

so $$\frac{1/n+1/n^{3}}{1+1/n^{3}}$$
that would make it $$\infty/1$$
divergent i guess Thanks

anonymous · Answer

1/infinity is how much Mathsofiya??

anonymous · Answer

1/infty is zero but infinity/1 = infty

anonymous · Answer

So, I think answer will be : 0/1 = 0

anonymous · Answer

what about $$a_{n}/b_{n}$$ ? my book multiplied n/1 to $$a_{n}$$

anonymous · Answer

I did not get what you said...
explain it more..

anonymous · Answer

The book calls it the limit comparison test 
$$\lim_{n\rightarrow\infty} \frac{a_{n}}{b_{n}}$$ and somehow they took an n/1 and multiplied it to $$ \frac{n^{2}+1}{n^{3} +1}$$

anonymous · Answer

Well, I have no knowledge about this.
Sorry, but I will search this and make you clear this one....

anonymous · Answer

the answer should be 0, you did it right o-o...