Find $\overline{\lim}\limits_{n\to\infty}\left(\frac{1}{n}-\frac{2}{n}+\frac{3}{n}-...+(-1)^{n-1}\frac{n}{n}\right)$ Find $\overline{\lim}\limits_{n\to\infty}\left(\frac{1}{n}-\frac{2}{n}+\frac{3}{n}-...+(-1)^{n-1}\frac{n}{n}\right)$
$\frac{1}{n}-\frac{2}{n}+\frac{3}{n}-...+(-1)^{n-1}\frac{n}{n}=\frac{1}{n}\sum\limits_{k=1}^{n}(-1)^{k-1}k$
How to evaluate sum $\sum\limits_{k=1}^{n}(-1)^{k-1}k$?
 A: Hint: The first few values of your sum are $1, -1, 2, -2, 3, -3, 4, -4, \ldots$. Try to derive a general formula for the sum depending on whether $n$ is even or odd. Then prove this formula via induction.
A: Consider the sum:
$$
S=\sum_{k=1}^{n} x^k=x\frac{1-x^n}{1-x}
$$
Take its derivative with respect to $x$:
$$
S'=\sum_{k=1}^{n}kx^{k-1}=\frac{nx^{n+1}-(n+1)x^n+1}{{(1-x)}^2}
$$
So for $x=-1$ you would get:
$$
\sum_{k=1}^{n}k{(-1)}^{k-1}=\frac{n{(-1)}^{n+1}-(n+1){(-1)}^{n}+1}{4}
$$
Now the limit would be:
$$
{lim}_{n\rightarrow\infty}\frac{S'}{n}={lim}_{n\rightarrow\infty} \frac{n{(-1)}^{n+1}-(n+1){(-1)}^n+1}{4n}\\
={lim}_{n\rightarrow\infty} \frac{{(-1)}^{n+1}(2n+1)+1}{4n}\\
={lim}_{n\rightarrow\infty} \frac{{(-1)}^{n+1}(2+\frac{1}{n})+\frac{1}{n}}{4}
$$
From the above we see that the limit doesnt exist as it would be either $\frac{1}{2}$ or $-\frac{1}{2}$ depending on wether $n$ is even or odd.  But as Michael pointed out, I was supposed to find the limit superior in which case the answer would be $\frac{1}{2}$.
