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 ? 

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:
$$ 1 = \lim_{n \to \infty} \left( n \frac{1}{n} \right)= \lim_{n \to \infty} \underbrace{\left( \frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n} \right)}_{n \ \text{times}} =\lim_{n \to \infty} \frac{1}{n}+\lim_{n \to \infty} \frac{1}{n} + \dots + \lim_{n \to \infty} \frac{1}{n} = 0 \neq 1 $$
