About two ways to compute a special limit and what is incorrect? I have an question in compute the following limit:
$$
\lim_{n\to \infty} \frac{1}{n} +\frac{2}{n} +...+\frac{n}{n} 
$$ 
I know that the answer is:
$$
\lim_{n\to \infty} \frac{1}{n} +\frac{2}{n} +...+\frac{n}{n}
=
\lim_{n\to \infty} \frac{\frac{n(n+1)}{2}}{n}
=
\infty
$$
but a way for compute this limit is:
$$
\lim_{n\to \infty} \frac{1}{n} +\frac{2}{n} +...+\frac{n}{n}
=
\lim_{n\to \infty}\frac{1}{n} + 
\lim_{n\to \infty}\frac{2}{n} +
\ldots +
\lim_{n\to \infty}\frac{n}{n}
=
\lim_{n\to \infty} 0+0+\ldots +1
=1
$$ 
what second way is not true?
 A: You can't distribute the limit over a number of summands that depends on $n$, as shown. The first solution is correct. You have
$${1\over n }\sum_{k=1}^n k = {n+1\over 2}$$
A: From my point, your method has some question. As is well known, if we add finte terms in the sense of limit, then you can plus every term after you calculate every limit, but if there are infinite terms, you can not do like this. 
A: You tried something like
$$
\lim_{n\to\infty} \sum_{k=1}^n \frac{k}{n} =
\lim_{N\to\infty} \sum_{k=1}^N \lim_{n\to\infty} \frac{k}{n}
$$
and tried to synchronize $N$ and $n$ somehow. 
A: In the OP, the conclusion that the limit 
$$\lim_{n\to \infty}\left(\frac1 n+\frac2 n+\cdots+\frac{n}{n}\right)=0+0+\cdots +1=1$$
is incorrect.  
We can see this more clearly by writing
$$\frac1 n+\frac2 n+\frac3n+\cdots+\frac{n-3}{n}+\frac{n-2}{n}+\frac{n-1}{n}+\frac{n}{n}$$
Now, we see that all of the last terms after the "dots" approach $1$ as $n\to \infty$.  
So, in order to approach this problem correctly, we need to write all of the terms as 
$$\frac1 n+\frac2 n+\frac3n+\frac4n+\cdots+\frac{n-3}{n}+\frac{n-2}{n}+\frac{n-1}{n}+\frac{n}{n}=\frac1n\sum_{k=1}^n\,k=\frac1n \frac{n(n+1)}{2}=\frac{(n+1)}{2}$$
and now clearly the limit is $\infty$.
