Douts about the limit criterion for series

Question

Well, I had to analyze convergence of the series $\frac{n+1}{n^2 +1}$, so I tried using the limit comparison test. ${a_n}$ would be $\frac{n+1}{n^2 +1}$ and I took $\frac{1}{x}$ as ${b_n}$. Then I computed the limit of $\frac{a_n}{b_n}$ when $n$ approaches $+ \infty$, and I got 1 as the result, which is different from zero so it means that both series diverge since $\frac{1}{x}$ diverges. However, if a take $\frac{1}{x^2}$ as ${b_n}$, the result of the limit is $+ \infty$ which is also different from zero, so ${a_n}$ must converge since $\frac{1}{x^2}$ converges?

Also, is there any difference when deciding which series is ${a_n}$ and which is ${b_n}$? I mean, if I had chosen $\frac{n+1}{n^2 +1}$ as ${b_n}$ and $\frac{1}{x}$ as ${a_n}$, would the result have been different?

Answer 1

Hint: Write your term in the form $$\frac{n(1+\frac{1}{n})}{n^2(1+\frac{1}{n^2})}=\frac{1}{n}\frac{1+\frac{1}{n}}{1+\frac{1}{n^2}}$$ so the searched limit is zero.

Answer 2

I suppose you meant to write $\;b_n=\cfrac1n\;$ , but then the limit is something else:

$$\frac{a_n}{b_n}=\frac{\frac{n+1}{n^2+1}}{\frac1n}=\frac{n^2+n}{n^2+1}\xrightarrow[n\to\infty]{}1$$

and then, since $\;\sum\limits_{n=1}^\infty\frac1n\;$ diverges, so does our series (of course, all series involved here are positive ones).