Applicability of L'Hôpital rule on infinite sum. $$\lim_{n \to \infty} \left( \frac{n}{n^2+1} + \frac{n}{n^2+2} + \frac{n}{n^2+3} + \space  ... \space + \frac{n}{n^2+n}\right) $$ 
As $n$ is not $\infty$ but tends to $\infty$ I can split the limit of sums into sum of limits. i.e.
$$\lim_{n \to \infty} \frac{n}{n^2+1} +\lim_{n \to \infty} \frac{n}{n^2+2} +\lim_{n \to \infty} \frac{n}{n^2+3} + \space  ... \space +\lim_{n \to \infty} \frac{n}{n^2+n} $$ 
Applying L'Hôpital rule.
$$\lim_{n \to \infty} \frac{1}{2n} +\lim_{n \to \infty} \frac{1}{2n} +\lim_{n \to \infty} \frac{1}{2n} + \space  ... \space +\lim_{n \to \infty} \frac{1}{2n} $$ 
$$= \lim_{n \to \infty} \left( \frac{1}{2n} + \frac{1}{2n} + \frac{1}{2n} + \space  ... \space + \frac{1}{2n} \right) = \lim_{n \to \infty} 
\frac{n}{2n} $$
$$ =\lim_{n \to \infty } \frac{1}{2} = \frac{1}{2}$$
Whereas applying Sandwich theorem with $g(x) \leq f(x) \leq h(x)$, where
$$g(x) = \frac{n}{n^2+n} + \frac{n}{n^2+n} + \frac{n}{n^2+n} + \space  ... \space + \frac{n}{n^2+n} = \frac{n^2}{n^2+n}$$
and
$$h(x) = \frac{n}{n^2+1} + \frac{n}{n^2+1} + \frac{n}{n^2+1} + \space  ... \space + \frac{n}{n^2+1} = \frac{n^2}{n^2+1 }$$
Yields
$$\lim_{n \to \infty} g(x) = \lim_{n \to \infty} h(x) = 1
\implies \lim_{n \to \infty} f(x) = 1
$$
Sandwich theorem is very intuitive to discard hence I suppose there is some issue with the application of L'Hôpital's rule.
Is there any special condition involved with converting limit of sums to sum of limits ? 
 A: Clearly the problem is way before L'Hopital's, splitting the limit into a sum of limits is not allowed (precisely because the number of terms tends to infinity, even though it is not infinite). For instance, consider:
$$ 1 = \lim_{n \to \infty} \left( n \frac{1}{n} \right)= \lim_{n \to \infty} \underbrace{\left( \frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n} \right)}_{n \ \text{times}} =\lim_{n \to \infty} \frac{1}{n}+\lim_{n \to \infty} \frac{1}{n} + \dots + \lim_{n \to \infty} \frac{1}{n} = 0 \neq 1 $$
A: Notice that if you allow splitting the limit into sum of limits you can arrange it to be any real number.
Indeed, take any $x \in \mathbb{R}$. For $k \in \{1, \ldots, n\}$ we have:
$$\displaystyle\lim_{n \to \infty} \frac{n}{n^2+k} = 0 = \lim_{n\to\infty} \frac{x}{n}$$
So we would obtain:
\begin{align}
\lim_{n \to \infty} \left( \frac{n}{n^2+1} + \cdots + \frac{n}{n^2+n}\right) &= \lim_{n \to \infty} \frac{n}{n^2+1} +\lim_{n \to \infty} \frac{n}{n^2+2} +\cdots + \lim_{n \to \infty} \frac{n}{n^2+n}\\
&= \lim_{n\to\infty} \frac{x}{n} + \lim_{n\to\infty} \frac{x}{n} + \cdots + \lim_{n\to\infty} \frac{x}{n}\\
&=\lim_{n\to\infty} \underbrace{\left(\frac{x}{n} + \cdots + \frac{x}n\right)}_{n \text{ times}}\\
&= \lim_{n\to\infty} x\\
&= x
\end{align}
This is clearly wrong as you have correctly shown that the limit is in fact $1$.
Also, it is not entirely clear what you mean by L'Hôpital's rule here. The following is not true in general:
$$\lim_{n\to\infty} \frac{f(n)}{g(n)} = \lim_{n\to\infty} \frac{f'(n)}{g'(n)}$$
For sequences an analogous (but not exactly the same) result holds.

Stolz–Cesàro Theorem:
If $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are two sequences such that $(b_n)_{n=1}^\infty$ is strictly increasing and unbounded, then:
$$\lim_{n\to\infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L \implies \lim_{n\to\infty} \frac{a_n}{b_n} = L$$

Applying this here we would obtain
$$\lim_{n\to\infty} \frac{n}{n^2 + k} = \lim_{n\to\infty} \frac{(n+1) - n}{(n+1)^2 + k - (n^2 + k)} = \lim_{n\to\infty}\frac{1}{2n + 1} = 0$$
which we already know, of course.
