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True. If a series converges then its terms tend to zero.

(a) Differentiating both sides of

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

gives

$$\sum_{n=0}^\infty n x^{n-1}=\frac{1}{(1-x)^2}$$

Now multiplying both sides by $x$ gives

$$\sum_{n=0}^\infty n x^n=\frac{x}{(1-x)^2}$$

as desired.

(b) Differentiating both sides of the conclusion of part (a) gives

$$\sum_{n=0}^\infty n^2 x^{n-1}=\frac{(1-x)^2-2x(x-1)}{(1-x)^4} =\frac{(1-x)(1-x+2x)}{(1-x)^4} =\frac{1+x}{(1-x)^3}$$

Now multiplying both sides by $x$ gives

$$\sum_{n=0}^\infty n^2 x^n=\frac{x(1+x)}{(1-x)^3}$$

We know that differentiation preserves the radius of convergence of power series. So this series has radius of convergence $1$ (the radius of convergence of the original geometric series). At $x=\pm1$ the series diverges by the divergence test. So the series converges for $-1<x<1$.

(a) We know that the radius of convergence $R$ obeys

$$\begin{aligned} \frac{1}{R} = \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty}\frac{n}{n+1}\ \frac{(n+1)a_{n+1}}{na_n} = 1 \frac{C}{C} = 1 \end{aligned}$$

because we are told that $\lim\limits_{n\rightarrow\infty} na_n=C$. So $R=1$.

(b) Just knowing that the radius of convergence is $1$, we know that the series converges for $|x|<1$ and diverges for $|x|>1$. That leaves $x\pm 1$.

When $x=+1$, the series reduces to $\sum\limits_{n=1}^\infty a_n$. We are told that $na_n$ decreases to $C>0$. So $a_n\ge \frac{C}{n}$. By the comparison test with the harmonic series $\sum\limits_{n=1}^\infty\frac{1}{n}$, which diverges by the $p$–test with $p=1$, our series diverges when $x=1$.

When $x=-1$, the series reduces to $\sum\limits_{n=1}^\infty (-1)^n a_n$. We are told that $na_n$ decreases to $C>0$. So $a_n>0$ and $a_n$ converges to $0$ as $n\rightarrow\infty$. Consequently $\sum\limits_{n=1}^\infty (-1)^n a_n$ converges by the alternating series test.

In conclusion $\sum\limits_{n=1}^\infty a_n x^n$ converges when $-1\le x< 1$.

First, we differentiate.

$$\begin{aligned} f(x)&=\displaystyle\sum_{n=0}^{\infty}A_n(x-c)^n\\ f'(x)&=\sum_{n=0}^{\infty}nA_n(x-c)^{n-1}\\ &=\sum_{n=1}^{\infty}nA_n(x-c)^{n-1}\\ f'(c)&=\sum_{n=1}^\infty nA_n\cdot 0^{n-1}\\ &=A_1\cdot 1 + 2A_2\cdot0+3A_3\cdot0+\cdots\\ &=A_1 \end{aligned}$$

So, if $A_1=0$, then $f'(c)=0$. That is, $f(x)$ has a critical point at $x=c$.

To determine the behaviour of this critical point, we use the second derivative test.

$$\begin{aligned} f'(x)&=\sum_{n=1}^{\infty}nA_n(x-c)^{n-1}\\ f''(x)&=\sum_{n=1}^{\infty}n(n-1)A_n(x-c)^{n-2}\\ &=\sum_{n=2}^{\infty}n(n-1)A_n(x-c)^{n-2}\\ f''(c)&=\sum_{n=2}^{\infty}n(n-1)A_n\cdot 0^{n-2}\\ &=2(1)A_2\cdot 0^0+3(2)A_3\cdot 0^1+4(3)A_4\cdot 0^2+\cdots\\ &=2A_2 \end{aligned}$$

Following the second derivative test, $x=c$ is the location of a local maximum if $A_2<0$, and it is the location of a local minimum if $A_2>0$. (If $A_2=0$, the critical point may or may not be a local extremum.)