Interchange of sum and limit in sequence algebra As you may know,
Let $ {a_n} $ and $ {b_n} $ be convergent sequence with limit L, M respectively, then the following is true
$ \lim_{n\to \infty} (a_n + b_n) = \lim_{n\to \infty} a_n +\lim_{n\to \infty} b_n = L+M$
In the following case
$$ \lim \frac{\sum_{k=1}^n k}{n^2} = \lim \frac{\frac{n(n+1)}{2}}{n^2} = \frac{1}{2}$$
is true but
$$ \lim \frac{\sum_{k=1}^n k}{n^2} = \lim \frac{1+2+3+\cdots+n}{n^2} = \lim \frac{1}{n^2} + \lim \frac{2}{n^2} + \cdots + \lim \frac{n}{n^2} = 0$$
I think this cannot be true because there are infinitely many terms.
However, let $ \alpha $ be a positive real number, then the following can be true
$$ \lim \frac{\sum_{k=1}^n k}{n^{2+\alpha}} = \lim \frac{1+2+3+\cdots+n}{n^{2+\alpha}} = \lim \frac{1}{n^{2+\alpha}} + \lim \frac{2}{n^{2+\alpha}} + \cdots + \lim \frac{n}{n^{2+\alpha}} = 0$$
Does this have something to do with uniform convergence?
When can I interchange the limit and sigma in limit operation of sequence?
 A: $\lim_{n\to\infty}\frac1{n^2}(1+2+\cdots+n)\ne\lim\frac1{n^2}+\cdots+\lim\frac n{n^2}$ becuase the number of terms varies with $n$.  You can only apply the sum of limits rule, if the number of terms stay fixed.
A: You can't interchange limit and sum when evaluating a limit of the form $\lim_{n\to \infty} \sum_{k = 1}^n a_k$ because the number of summands of the sequence $\sum_{k = 1}^n a_k$ depends on $n$.
A: Unfortunately what you've written makes little sense. You aren't actually interchanging sum and limit but rather somewhat abusing the definition of a sum as Tim implies in his answer. 
However, you can in some sense make an interpretation of what you've written as a Riemann integral. For example:
$$\frac{1}{2}=\int_0^1 xdx = \lim_{n\rightarrow\infty} \frac{1}{n} \sum_{k=1}^n k/n,$$
which comes from:
$$\int_0^1 f(x)dx = \lim_{n\rightarrow\infty} \sum_{k=1}^n f(k/n)\frac{1}{n}.$$
So if you squint, your interpretation is basically that of a Riemann Integral: you are adding up pieces of height $f(k/n)$ and width $1/n$ which have smaller and smaller area. It is wrong to say that you can pass the limit to each term because then you really do get zero, which is wrong. 
