In your Limits?

$\begin{aligned} \lim_{n \to \infty} \sum_{k=1}^n \frac{k}{n^2+k} & = \lim_{n \to \infty} \sum_{k=1}^n \frac{\frac{k}{n^2}}{\color{#3D99F6}{1+\frac{k}{n^2}}} & \quad \quad \small \color{#3D99F6}{\text{Taylor's expansion on } \frac{1}{1+\frac{k}{n^2}}} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac{k}{n^2} \color{#3D99F6} {\left(1-\frac{k}{n^2} + \frac{k^2}{n^4} - \frac{k^3}{n^6} + ... \right)} & \quad \quad \small \color{#3D99F6}{\text{As } n \to \infty, \left(1-\frac{k}{n^2} + \frac{k^2}{n^4} - ... \right) \to 1} \\ & = \lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n k \\ & = \lim_{n \to \infty} \frac{1}{n^2} \left(\frac{n(n+1)}{2}\right) \\ & = \lim_{n \to \infty} \frac{1+\frac{1}{n}}{2} \\ & = \frac{1}{2} = \boxed{0.5} \end{aligned}$

We use the squeeze theorem to compute this limit.

As a lower bound, since $k \le n$ , we have:

$\displaystyle\begin{aligned} \sum_{k=1}^n \frac{k}{n^2+k} &\ge \sum_{k=1}^n \frac{k}{n^2+n} \\ &= \frac{\sum_{k=1}^n k}{n(n+1)} \\ &= \frac{\frac{n(n+1)}{2}}{n(n+1)} \\ &= \frac{1}{2} \end{aligned}$

As an upper bound, since $k \ge 0$ , we have: $\displaystyle\begin{aligned} \sum_{k=1}^n \frac{k}{n^2+k} &\le \sum_{k=1}^n \frac{k}{n^2} \\ &= \frac{\sum_{k=1}^n k}{n^2} \\ &= \frac{\frac{n(n+1)}{2}}{n^2} \\ &= \frac{n+1}{2n} \\ &= \frac{1}{2} + \frac{1}{2n} \end{aligned}$

The lower bound is constant at $\frac{1}{2}$ . The upper bound tends to $\frac{1}{2}$ as $n$ goes to infinity. By the squeeze theorem, the number between them, $\displaystyle\sum_{k=1}^n \frac{k}{n^2+k}$ , also tends to $\frac{1}{2} = \boxed{0.5}$ .

$\displaystyle\lim_{n\rightarrow \infty} \displaystyle\sum_{k=1}^n \dfrac{k}{n^2 + k} = \displaystyle\lim_{n\rightarrow \infty} \displaystyle\sum_{k=1}^n \dfrac{1}{n} \left(\dfrac{\dfrac{k}{n}}{1 + \dfrac{k}{n^2}}\right)$
We will put $\dfrac{k}{n} = x$ and $\dfrac{1}{n} \rightarrow \text{dx}$ as ( n $\rightarrow \infty$ ).
Therefore,given sum will convert into integral with limits 0 to 1.
$\displaystyle\int_{0}^{1} x\text{dx} \quad \quad (\dfrac{k}{n^2} \rightarrow 0)$ .

$\Rightarrow \dfrac{x^2}{2} | _0^1 = \color{#3D99F6}{\boxed{0.5}}$