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We can find $f(x)$ by differentiating its integral, or antidifferentiating its derivative. In the latter case, we'll have to solve for the arbitrary constant of integration; in the former case, we do not. (Remember that many different functions have the same derivative, but a single function has only one derivative.) To avoid the necessity of finding the arbitrary constant, we can ignore the given equation for $f'(x)$, which makes the problem much simpler. This is the method used in Solution 1.

• Solution 1 Using the Fundamental Theorem of Calculus Part 1:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \left\{\int_5^x f(t)\mathrm{d}t \right\}&=f(x)\\ \text{So,}\qquad f(x)&=\frac{\mathrm{d}}{\mathrm{d}x} \left\{3x+\displaystyle\sum_{n=0}^\infty \frac{(x-1)^{n+1}}{n(n+1)^2} \right\}\\ &=3+\displaystyle\sum_{n=1}^\infty \frac{(n+1)(x-1)^{n}}{n(n+1)^2}\\ &=3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)} \end{aligned}$$

• Solution 2 Suppose we had used $f'(x)$ instead. We would antidifferentiate to find:

$$\begin{aligned} f(x)&=\int \left( \sum_{n=0}^\infty \frac{(x-1)^{n}}{n+2} \right)\mathrm{d}x\\ &=\left(\sum_{n=0}^\infty \frac{(x-1)^{n+1}}{(n+1)(n+2)}\right)+C\\ &=\left(\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)}\right)+C \end{aligned}$$

Notice $f(1)=0+C$. So, to find $C$, we must find $f(1)$. We can't get that information from $f'(x)$, so our only option is to consider the given formula for $\int_5^x f(t)\mathrm{d}t$. Using the Fundamental Theorem of Calculus Part 1:

$$\begin{aligned} f(1)&=\left.\frac{\mathrm{d}}{\mathrm{d}x}\left\{\int_5^x f(t)\mathrm{d}t \right\}\right|_{x=1}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d}x}\left\{ 3x+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n+1}}{n(n+1)^2} \right\}\right|_{x=1}\\ &=\left[ 3+\displaystyle\sum_{n=1}^\infty \frac{(n+1)(x-1)^{n}}{n(n+1)^2} \right]_{x=1}\\ &=\left[ 3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)} \right]_{x=1}\\ &= 3+\displaystyle\sum_{n=1}^\infty \frac{0^{n}}{n(n+1)}\\ &=3 \end{aligned}$$

So, $f(x)=3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)}$.

Note that in Solution 2, we did the same calculation as Solution 1, and more.

We'll start by applying the ratio test twice,

• once for the series whose $n^{\rm th}$ term is $a_n=\frac{x^n}{3^{2n}\log n}$, and
• once for the series whose $n^{\rm th}$ term is $a_n=\big|\frac{x^n}{3^{2n}\log n}\big|$.

For both series, i.e. for both $a_n$'s,

$$\begin{aligned} \lim_{n\to\infty} \bigg| \frac{ a_{n+1} }{ a_n } \bigg| &= \lim_{n\to\infty} \bigg| \frac{x^{n+1}}{3^{2(n+1)}\log(n+1)} \ \frac{3^{2n}\log n}{x^n} \bigg| \\ &= \lim_{n\to\infty} \bigg| \frac{x\log n}{3^2\log(n+1)} \bigg| =\lim_{n\to\infty} \bigg| \frac{x\log n}{3^2[\log(n)+\log(1+1/n)]} \bigg| \\ &=\lim_{n\to\infty} \bigg| \frac{x}{3^2[1+\log(1+1/n)/\log(n)]} \bigg| \\ &=\frac{|x|}{9} \end{aligned}$$

Therefore, by the ratio test, when $|x|<9$, the series $\sum_{n=2}^\infty \big|\frac{x^n}{3^{2n}\log n}\big|$ converges and the series $\sum_{n=2}^\infty \frac{x^n}{3^{2n}\log n}$ converges absolutely, and when $|x|> 9$ the series $\sum_{n=2}^\infty \frac{x^n}{3^{2n}\log n}$ diverges. That just leaves $x=\pm 9$.

For $x=-9$, $\displaystyle\sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} =\sum_{n=2}^\infty\frac{(-1)^n}{\log n}$ which converges by the alternating series test.

For $x=+9$, $\displaystyle\sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} =\sum_{n=2}^\infty\frac{1}{\log n}$ which is the same series as $\displaystyle\sum_{n=2}^\infty\Big|\frac{(-1)^n}{\log n}\Big|$. We shall shortly show that $n\ge\log n$, and hence $\frac{1}{\log n}\ge \frac{1}{n}$ for all $n\ge 1$. This implies that the series $\displaystyle\sum_{n=2}^\infty\frac{1}{\log n}$ diverges by comparison with the divergent series $\displaystyle\sum_{n=2}^\infty\frac{1}{n^p}\bigg|_{p=1}$. This yelds both divergence for $x=9$ and also the failure of absolute convergence for $x=-9$.

Finally, we show that $n-\log n > 0$, for all $n\ge 1$. Set $f(x)=x-\log x$. Then $f(1)=1>0$ and

$$f'(x) = 1 -\frac{1}{x} \ge 0\qquad\text{for all }x\ge 1$$

So $f(x)$ is (strictly) positive when $x=1$ and is increasing for all $x\ge 1$. So $f(x)$ is (strictly) positive for all $x\ge 1$.

(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.)

We recognize $\displaystyle \sum_{n=3}^\infty \frac{n}{5^{n-1}}$ as $f(x)=\displaystyle \sum_{n=3}^\infty n\cdot x^{n-1}$, evaluated at $x=\dfrac15$. We should figure out what $f(x)$ is in equation form (as opposed to power series form). Notice that this looks similar to the derivative of the geometric series $\displaystyle\sum x^n$.

$$\begin{aligned} \frac{1}{1-x}&=\sum_{n=0}^\infty x^n \qquad \text{when $|x|<1$}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{1-x}\right\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sum_{n=0}^\infty x^n\right\}\\ \frac{1}{(1-x)^2}&=\sum_{n=1}^\infty nx^{n-1}\\ &= 1x^0 + 2x^1+\sum_{n=3}^\infty nx^{n-1}\\ &= 1 + 2x+\sum_{n=3}^\infty nx^{n-1}\\ \text{So,}\qquad \frac{1}{(1-x)^2}-1-2x&=\sum_{n=3}^\infty nx^{n-1}\\ \text{Setting $x=\dfrac15$:}\qquad \frac{1}{(1-1/5)^2}-1-\frac25&=\sum_{n=3}^\infty n\left(\frac15\right)^{n-1}\\ \left(\frac{5}{4} \right)^2-1-\frac25&=\sum_{n=3}^\infty \frac{n}{5^{n-1}}\\ \end{aligned}$$

So, our series evaluates to $\displaystyle\frac{25}{16}-1-\frac{2}{5}=\frac{13}{80}$.