Show that the series does not converge uniformly on the interval $I=(-r,0]$ Let $x\in\mathbb{R}$ and consider the power series: $$\sum_{n=0}^\infty (-1)^n(n^2+3n+2)x^n.$$
a) Find the radius of convergence $r$ for the series.  

b) Show that the series is the Taylor series for $f(x)=\frac{2}{(1+x)^3}$ in $x=0$.  

c) Show that the series does not converge uniformly on the interval $I=(-r,0]$. (You can, without proof, use the fact that the series is convergent towards $f$ on the interval of convergence.)  

...In a) I've found the radius of convergence to be $r=1$. Can anybody help me with question c)?
To show that a serie is uniformly convergent, you generally show that

$$R_N(x) = \sum_{n=N}^\infty (-1)^n(n^2+3n+2)x^n \xrightarrow[{N\to\infty}]{} 0 \qquad \text{uniformly}$$ 

Namely, because for $x &lt; 0$ the sum is always positive, if we note $R_N(x)$ this sum, that $\forall \epsilon, \exists N_0 &gt; 0, \forall N &gt; N_0, \forall x \in ]-1;0], R_N(x) \leq \epsilon$. But here we are going to show the opposite, namely, $\exists \epsilon, \forall N_0 &gt; 0, \exists N &gt; N_0, \exists x \in ]-1;0], R_N(x) &gt; \epsilon$. which can be written $\exists \epsilon, \forall N &gt; 0, \exists x \in ]-1;0], R_N(x) &gt; \epsilon$. Any $\epsilon$ can work, so:

Fixing $A &gt; 0$, no matter how big you choose $N$, you will always be able to find $x \in ]-1;0]$ such that it is bigger than $A$ (because it still tends to infinity). This shows the opposite of uniformly convergence, hence it is not uniformly convergent.

To show this:
Fix $\epsilon = 1$, $N &gt; 0$. The function $x \mapsto R_N(x)$ is decreasing with $R_N(x) \xrightarrow[{x\to -1}]{} +\infty$. So, by definition we can find a $x$ such that $R_N(x) &gt; \epsilon$. 

And to show that $R_N(x) \xrightarrow[{x\to -1}]{} +\infty$, it is sufficient to show, that is not upper bounded (by monotonicity). If it was upper bounded by $K$. Then the sum from $N$ to $M$ would be upperd bounded by $K$ (for all $x$). So you make $x$ tend to $-1$ and then $M$ tend to infinity and you shouw that $R_N(-1) &lt; +\infty$ which is not true.
For the question c : it's simple , you see it's an alternating series, meaning it's something-something+something-something+something ....
I'll call $U_n=(-1)^n(n^2+3n+2)x^n$
In order for the series to converge as an alternating series,
$$\lim_{n\to\infty}\lvert\frac{U_{n+1}}{U_n}\rvert&lt;1$$
As well as having $\lim_{n\to\infty}U_n=0$

Let's first evaluate $\frac{U_{n+1}}{U_n}$ :
$$\frac{U_{n+1}}{U_n}=\frac{(-1)^{n+1}((n+1)^2+3(n+1)+2)x^{n+1}}{(-1)^n(n^2+3n+2)x^n}=-x\times\frac{n^2+5n+6}{n^2+3n+2}$$
And so :
$$\lim_{n\to\infty}\frac{U_{n+1}}{U_n}=-x\times\lim_{n\to\infty}\frac{n^2+5n+6}{n^2+3n+2}$$

we see that in the second limit the highest powers are 2, and the coefficients of both are equal to $1$ so the limit is equal :
$$\lim_{n\to\infty}\frac{U_{n+1}}{U_n}=-x$$
since $x\in(-1,0]$, then $0\leq\lim_{n\to\infty}\frac{U_{n+1}}{U_n}&lt;1$ And so the first test is aqcuired and returned with success. The second tests now consits of proving $\lim_{n\to\infty}U_n=0$ so let's do that :
$$\lim_{n\to\infty}U_n=\lim_{n\to\infty}(-1)^n(n²+3n+2)x^n$$
We have $x^n$ an exponential function thus we treat it first. We know that $x\in(-1,0]$ so $x^n$ is a decreasing function meaning that $\lim_{n\to\infty}x^n=0$ and so $\lim_{n\to\infty}U_n=0$ and the second test is passed.
That means that the series $\sum_{n=0}^{\infty}(-1)^n(n²+3n+2)x^n$ for $I=(-r,0]$ converges.

I hope that helped, it's called the Alternating series test.