I have to evaluate $\lim_{n \to \infty}n/(n^2+1) + n/(n^2+2)+n/(n^2+3)+....+n/(n^2+n)$
Now, for the rth term: $\lim_{n \to \infty}n/(n^2+r) = 1/(n+(r/n)) = 0.$ Since this is true for any term, and limits are distributive over addition, the limit should be zero. But this is wrong. Why?
hint
$$\frac {n^2}{n^2+n}\le\sum_{k=1}^n \frac {n}{n^2+k} \le \frac {n^2}{n^2+1}$$
Squeeze.
First of all let's get the terminology correct: "distribute" refers to a property of two binary operations such as addition and multiplication, $$a(b+c)=ab+ac\ .$$ Limits do not fall into this category. You might say that "limits preserve addition" or "limits respect addition" or "the limit of a sum is the sum of the limits".
What this means, more formally, is that $$\lim(a_n+b_n)=\lim a_n+\lim b_n\ .$$ To put it even more carefully,
if $\lim a_n$ and $\lim b_n$ both exist, then $\lim(a_n+b_n)$ exists, and $$\lim(a_n+b_n)=\lim a_n+\lim b_n\ .$$
You can easily extend this to the limit of a sum of three terms, or four, or any fixed number. However, in your example you have a sum of $n$ terms, which is not a fixed number - it increases as $n$ increases. So you have a situation in which your terms are getting smaller and smaller, but you are adding up more and more of them: so it should make sense intuitively that the sum might not tend to zero.
An example which works in much the same way as yours but with easier calculations: $$\eqalign{ \lim_{n\to\infty}\Bigl(\frac{n}{n^2}+\frac{2n}{2n^2}+\frac{3n}{3n^2}+\cdots +\frac{n.n}{n.n^2}\Bigr) &=\lim_{n\to\infty}\Bigl(\frac{1}{n}+\frac{1}{n}+\frac{1}{n}+\cdots +\frac{1}{n}\Bigr)\cr &=\lim_{n\to\infty}1\cr &=1\ ,\cr}$$ even though each term inside the brackets tends to zero as $n\to\infty$.
\infty. – David Jun 21 '17 at 03:34