Interpreting the ratio test for $\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k} $ We want to find the limit of this.
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k} $$
I would have done it as follows:
$$\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| = \lim_{n\to\infty}\frac{\frac{n+1}{(n+1)^2+n}}{\frac{n}{n^2+n}} =\lim_{n\to\infty} \frac{n+1}{(n+1)^2+n} \cdot \frac{n^2+n}{n} = \lim_{n\to\infty}\frac{n^3+2n^2+n}{n^3+3n^2+n} \\
=\lim_{n\to\infty} \frac{n^3 \cdot \bigl(1+\frac{2}{n} + \frac{1}{n^2} \bigr)}{n^3\cdot\big(1+\frac{3}{n} + \frac{1}{n^2} \bigr)} = \frac{1}{1} = 1$$
According to the ratio test, the series converges if $\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| <1$ and it diverges if it's $> 1$.
But since we get  $1$ here, the series converges towards that value, no? But according to the ratio test, we can't make a statement about the series of the limit of $\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| = 1$
What am I misunderstanding here?

If you are familiar with harmonic numbers
$$a_n= \sum_{k=1}^n \frac{n}{n^2+k}=n \left(H_{n^2+n}-H_{n^2}\right)$$
Using the asymptotics
$$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ and continuing with Taylor series
$$a_n=1-\frac{1}{2 n}-\frac{1}{6 n^2}+\frac{1}{4
   n^3}+O\left(\frac{1}{n^4}\right)$$
Use it for $n=10$ : the exact result is
$$a_{10}=\frac{11210403701434961}{11818204429243212}=0.948571$$ while the truncated series gives
$$\frac{11383}{12000}=0.948583$$
