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?
 A: No, ratio test is used when
$$\lim_{n\to\infty} \sum_{k=1}^n a_k ~~~~~~~~~~\lim_{k \to \infty} \bigg| \frac{a_{k+1}}{a_k}\bigg|$$
But here you have
$$\lim_{n\to\infty} \sum_{k=1}^n a_k(n)$$
A: Will you allow me a proof without the ratio test which I think you
didn't use it in the correct way. We are interested in the limit of the
sum $s_{n}=n[\dfrac{1}{n^{2}+1}+\dfrac{1}{n^{2}+2}+....+\dfrac{1}{n^{2}+n}]$.
Notice that $\dfrac{1}{n^{2}+1}<\dfrac{1}{n^{2}}$ and likewise .... till
$\dfrac{1}{n^{2}+n}<\dfrac{1}{n^{2}}$. Hence the sum in brackets is less or equal to:
$\dfrac{n}{n^{2}}$=$\dfrac{1}{n}$.
Therefore $s_{n}\leq n\dfrac{1}{n}=1$. Also the sum in brackets is $\geq$ than $\dfrac{n}{n^{2}+n}$, which is $\dfrac{1}{n+1}$. Multiplied by $n$  we get $s_{n}\geq\,\dfrac{n}{n+1}$.Thus the sequence $s_{n}$ is such that $\dfrac{n}{n+1}\,\leq\,s_{n}\leq\,1$ and certainly converges to $1$!! Therefore the series converges to $1$.!!
A: 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
$$\frac{11383}{12000}=0.948583$$
A: Unfortunately, the Ratio Test doesn't work for that kind of sum you have.
Recall for $n > 1$ and any decreasing function $f: \mathbb{R} \to \mathbb{R}$ that
$$\int_{1}^{n+1}f(x)dx \leq \sum_{k=1}^{n}f(k) \leq \int_{0}^{n}f(x)dx.$$
Let $f(x) = \frac{n}{n^2+x}.$ Then after some integrating, we get
$$\int_1^{n+1}f(x)dx = n\ln{\left(1+\frac{n}{n^2+1}\right)}$$
and
$$\int_0^nf(x)dx = n\ln{\left(1+\frac{1}{n}\right)}.$$
Take the limit as $n$ goes to infinity and apply the Squeeze Theorem. I think you can take it from there. Your final answer should be $1$.
