I am asked to evaluate the following question:
$$\lim_{n\to\infty} \left(\frac{1}{1-n^2}+\frac{2}{1-n^2}+...+\frac{n}{1-n^2}\right)$$
I did the following:
Using the property: Limit of a sum is the sum of limits, I take on each term separately.
$$\lim_{n\to\infty}(\frac{1}{1-n^2}) = \lim_{n\to\infty}(\frac{\frac{1}{n^2}}{\frac{1}{n^2}-1}) = \frac{0}{-1} = 0.$$
Similarly each term evaluates to $0$ despite the dependence on n:
$$\lim_{n\to\infty}(\frac{n}{1-n^2}) = \lim_{n\to\infty}(\frac{\frac{n}{n^2}}{\frac{1}{n^2}-1}) = 0.$$
I checked the answer and it said -$\frac{1}{2}.$ I checked the complete solution:
$$\lim_{n\to\infty}(\frac{1}{1-n^2}+\frac{2}{1-n^2}+...+\frac{n}{1-n^2}) = \lim_{n\to\infty}(\frac{1+2+...+n}{1-n^2}) = \lim_{n\to\infty}(\frac{\frac{n(n-1)}{2}}{1-n^2}) = -\frac12.$$
Obviously, only one answer is ryt [hoping that the limit of an expression is a uniques number]. Both the deductions seem quite logical to me. The only plausible mistake i have noticed is $1+2+...+n = \frac{n(n-1)}{2}$ is only true for $n$ $\in$ $\Bbb Z$.
Solve the paradox.
The property you cite, "Limit of a sum is the sum of limits", does not apply here. That is for a sum of a fixed, finite number of terms. Formally, $$\lim_{n\to\infty}\sum_{k=1}^Ma_k(n)=\sum_{k=1}^M\lim_{n\to\infty}a_k(n)$$ The important part is $M$ is a fixed constant.
Your situation is $$\lim_{n\to\infty}\sum_{k=1}^na_k(n)$$
Note that even if you tried to take the limit to the terms inside, you would have $$\sum_{k=1}^n\lim_{n\to\infty}a_k(n)$$ which makes no sense. Since $n$ is the index variable for the limit, it has no meaning outside the limit, and could not appear as the sum's upper index.