Tricky Algebra Question I

Let's check for $x=-1$ :

$\frac{n(-1)^n}{(n+1)(-2+1)^n}=\frac{n(-1)^n}{(n+1)(-1)^n}$

We see that we can cancel out the $(-1)^n$ on both sides since it will never equal zero:

$\frac{n}{n+1}$

If we compute the limit of this expression as $n$ goes to infinity, we get:

$\displaystyle\lim_{n \rightarrow \infty} \frac{n}{n+1}=\displaystyle\lim_{n \rightarrow \infty} \frac{n}{n}=1$

Now,

$\displaystyle\sum_{n=0}^\infty 1= 1 \times \infty = \infty$

Therefore, when $x=-1$ , the series is divergent.