Comparison test to determine convergence

[masonm127]

Use a comparison test to determine whether the series $\sum$ (n+1)/(n$^{2}$+n+1) diverges or converges.

I started out by simplifying the series to 1/n+1 and then from there I compared it to 1/n, which converges. 1/n is greater than 1/n+1 so based on the comparison test, the original series should also converge, is this correct? I also tried a limit comparison test and got n/n+1 which equals 2 which would mean that both Ʃa and Ʃb converge. I am kind of shady on my series and and getting very confused with this question.
thanks!

 

[Mathitalian]

You could use this inequality $n^2+n+1\le n^2+2n+1= (n+1)^2\quad\forall n\in\mathbb{N}$ so:

$\frac{1}{(n+1)^2}\le \frac{1}{n^2+n+1}$

multiplying both sides for $n+1$, we have:

$\frac{1}{n+1}\le \frac{n+1}{n^2+n+1}$

So

$\sum_{n=0}^\infty\frac{1}{n+1}\le \sum_{n=0}^\infty\frac{n+1}{n^2+n+1}$

but
$$\sum_{n=0}^{\infty}\frac{1}{n+1}=\sum_{m=1}^{\infty}\frac{1}{m}=\infty$$

therefore $\sum_{n=0}^\infty\frac{n+1}{n^2+n+1}=\infty$

 

[HallsofIvy]

First, what do you mean by "simplified the series to 1/n+ 1". "Simplifying" normally reducing to something equal by canceling, say. That is not the case here. Second, the series $\sum (1/n)$ does NOT converge.