Active Calculus

1. We use the Ratio Test with \(a_k = \frac{|x-1|^k}{3k}\text{:}\)

   \begin{align*} \lim_{k \to \infty} \frac{ \frac{|x-1|^{k+1}}{3(k+1)} }{ \frac{|x-1|^k}{3k} } \amp = \lim_{k \to \infty} \frac{3k|x-1|^{k+1}}{3( k+1)|x-1|^{k}}\\ \amp = |x-1| \lim_{k \to \infty} \frac{k}{k+1}\\ \amp = |x-1|\text{.} \end{align*}

   So the power series \(\sum_{k=1}^{\infty} \frac{(x-1)^k}{3k}\) converges absolutely when \(|x-1| \lt 1\) or when \(0 \lt x \lt 2\) and diverges outside this interval. To completely determine the interval of convergence, we need to check what happens at the endpoints of this interval.

   • When \(x=0\) our power series is \(\sum_{k=1}^{\infty} \frac{(-1)^k}{3k}\) which is just a scalar multiple of the alternating harmonic series and so converges.

   • When \(x=2\) our power series is \(\sum_{k=1}^{\infty} \frac{1}{3k}\) which is just a scalar multiple of the harmonic series and so diverges.

   Therefore, the interval of convergence of the power series \(\sum_{k=1}^{\infty} \frac{(x-1)^k}{3k}\) is \([0,2)\text{.}\)

2. We use the Ratio Test with \(a_k = k|x|^k\text{:}\)

   \begin{align*} \lim_{k \to \infty} \frac{ (k+1)|x|^{k+1} }{ k|x|^k } \amp = |x|\lim_{k \to \infty} \frac{k+1}{k}\\ \amp = |x|\text{.} \end{align*}

   So the power series \(\sum_{k=1}^{\infty} kx^k\) converges absolutely when \(|x| \lt 1\) or when \(-1 \lt x \lt 1\) and diverges outside this interval. To completely determine the interval of convergence, we need to check what happens at the endpoints of this interval.

   • When \(x=-1\) our power series is \(\sum_{k=1}^{\infty} (-1)^k k\text{.}\) Since \(k \to \infty\) as \(k \to infty\text{,}\) this series diverges by the Divergence Test.

   • When \(x=1\) our power series is \(\sum_{k=1}^{\infty} k\) which again diverges by the Divergence Test.

   Therefore, the interval of convergence of the power series \(\sum_{k=1}^{\infty} kx^k\) is \((-1,1)\text{.}\)

3. We use the Ratio Test with \(a_k = \frac{k^2|x+1|^k}{4^k}\text{:}\)

   \begin{align*} \lim_{k \to \infty} \frac{ \frac{(k+1)^2|x+1|^{k+1}}{4^{k+1}} }{ \frac{k^2|x+1|^k}{4^k} } \amp = \lim_{k \to \infty} \frac{4^k(k+1)^2|x+1|^{k+1}}{4^{k+1}k^2|x+1|^k}\\ \amp = \frac{1}{4}|x+1| \lim_{k \to \infty} \left(\frac{k+1}{k}\right)^2\\ \amp = \frac{1}{4}|x+1|\text{.} \end{align*}

   So the power series \(\sum_{k=1}^{\infty} \frac{k^2(x+1)^k}{4^k}\) converges absolutely when \(\frac{1}{4}|x+1| \lt 1\) or when \(-5 \lt x \lt 3\) and diverges outside this interval. To completely determine the interval of convergence, we need to check what happens at the endpoints of this interval.

   • When \(x=-5\) our power series is \(\sum_{k=1}^{\infty} (-1)^k k^2\text{.}\) Since \(k^2 \to \infty\) as \(k \to \infty\text{,}\) this series diverges by the Divergence Test.

   • When \(x=3\) our power series is \(\sum_{k=1}^{\infty} k^2\text{,}\) which again diverges by the Divergence Test.

   Therefore, the interval of convergence of the power series \(\sum_{k=1}^{\infty} \frac{k^2(x+1)^k}{4^k}\) is \((-5,3)\text{.}\)

4. We use the Ratio Test with \(a_k = \frac{|x|^k}{(2k)!}\text{:}\)

   \begin{align*} \lim_{k \to \infty} \frac{ \frac{|x|^{k+1}}{(2(k+1))!} }{ \frac{|x|^k}{(2k)!} } \amp = \lim_{k \to \infty} |x|\frac{(2k)!}{(2(k+1))!}\\ \amp = |x| \lim_{k \to \infty} \frac{1}{(2k+2)(2k+1)}\\ \amp = 0\text{.} \end{align*}

   So the power series \(\sum_{k=1}^{\infty} \frac{x^k}{(2k)!}\) converges absolutely on the interval \((-\infty, \infty)\text{.}\)

5. We use the Ratio Test with \(a_k = k!|x|^k\text{:}\)

   \begin{align*} \lim_{k \to \infty} \frac{ (k+1)!|x|^{k+1} }{ k!|x|^k} \amp = \lim_{k \to \infty} |x|(k+1)\\ \amp = \infty \end{align*}

   unless \(x=0\text{.}\) So the interval of convergence of the power series \(\sum_{k=1}^{\infty} \frac{x^k}{k!}\) is \(\{0\}\text{.}\)

1. Recall that

   \begin{equation*} g(x) = \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \end{equation*}

   for \(-1 \lt x \lt 1\text{.}\)

2. Substituting \(-x^2\) for \(x\) in the power series expansion for \(g(x)\) gives

   \begin{align*} f(x) \amp = g(-x^2)\\ \amp = \frac{1}{1-(-x^2)}\\ \amp = \sum_{k=0}^{\infty} \left(-x^2\right)^k\\ \amp = \sum_{k=0}^{\infty} (-1)^k x^{2k}\text{.} \end{align*}

3. This power series expansion for \(f(x)\) will be valid as long as \(-1 \lt (-x)^2 \lt 1\) or for \(-1 \lt x \lt 1\text{.}\)

1. Note that

   \begin{equation*} f'(x) = - x + \frac{x^3}{3!} - \frac{x^5}{5!} + \cdots + (-1)^k \frac{x^{k-1}}{(k-1)!} + \cdots\text{.} \end{equation*}

2. We recognize \(f(x)\) as \(\cos(x)\) and \(f'(x)\) as \(-\sin(x)\text{.}\) This gives the known differentiation formula

   \begin{equation*} \frac{d}{dx} \cos(x) = -\sin(x)\text{.} \end{equation*}

3. We see that

   \begin{equation*} f''(x) = - 1 + \frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\text{,} \end{equation*}

   which is the series expansion for \(-\cos(x)\) as expected

Using the Taylor series

for \(-1 \lt x \lt 1\) we have that

Integrating the left and right sides of this last equation gives us

for \(-1 \lt x \lt 1\text{.}\) Since \(\ln(1) = 0\) we have that

and so \(C = 0\text{.}\) We conclude that

for \(-1 \lt x \lt 1\text{.}\)