Power Series practice Problems with solutions

Solution to part 1:

Let \( a_n = \dfrac{n+1}{3^n} x^n \). Compute:

\[ \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(n+2)\,x^{n+1}}{3^{n+1}} \cdot \frac{3^n}{(n+1)\,x^n}\right| = \frac{n+2}{3(n+1)}\,|x| \to \frac{|x|}{3} \quad \text{as } n\to\infty. \]

The Ratio Test requires the limit to be less than 1, so \( \dfrac{|x|}{3} < 1 \), giving \( \boxed{R = 3} \).

Solution to part 2:

At \( x = 3 \): the series becomes \( \displaystyle\sum_{n=0}^{\infty}(n+1) \), which diverges (terms do not tend to zero).

At \( x = -3 \): the series becomes \( \displaystyle\sum_{n=0}^{\infty}(-1)^n(n+1) \), which also diverges for the same reason.

Therefore the interval of convergence is \( (-3,\, 3) \).

Solution to part 1:
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{1}{(n+1)\cdot 5^{n+1}}\cdot\frac{n\cdot 5^n}{1}\cdot|x-2| = \frac{n}{5(n+1)}|x-2| \to \frac{|x-2|}{5}. \]

Convergence requires \( |x – 2| < 5 \), so \( R = 5 \). The open interval is \( (-3,\, 7) \).

Solution to part 2:

At \( x = 7 \) (so \( x – 2 = 5 \)):

\[ \sum_{n=1}^{\infty} \frac{(-1)^n}{n\cdot 5^n}\cdot 5^n = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}. \]

This is the alternating harmonic series, which converges by the Alternating Series Test.

At \( x = -3 \) (so \( x – 2 = -5 \)):

\[ \sum_{n=1}^{\infty} \frac{(-1)^n}{n\cdot 5^n}\cdot(-5)^n = \sum_{n=1}^{\infty} \frac{(-1)^n(-1)^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n}. \]

This is the harmonic series, which diverges.

The interval of convergence is \( (-3,\, 7] \).

Solution to part 1:

Let \( a_n = \left(\dfrac{n}{n+1}\right)^{n^2} x^n \). Apply the Root Test:

\[ \sqrt[n]{|a_n|} = \left(\frac{n}{n+1}\right)^n |x| = \frac{|x|}{\left(1 + \frac{1}{n}\right)^n} \to \frac{|x|}{e}. \]

Convergence requires \( \dfrac{|x|}{e} < 1 \), giving \( |x| < e \). Therefore \( \boxed{R = e} \).

Solution to part 2:

The Ratio Test would require computing \( \left(\dfrac{n}{n+1}\right)^{n^2} / \left(\dfrac{n+1}{n+2}\right)^{(n+1)^2} \), an expression with unwieldy exponent differences. The Root Test handles the \( n^2 \) exponent cleanly by reducing it to an \( n \)-th power, immediately invoking the classic limit for \( e \). This illustrates that the Root Test is superior when coefficients involve \( n \)-th powers or super-exponential decay.

Solution to part 1:

Write \( f(x) = \dfrac{4}{1-(-2x^2)} \). The substitution is \( u = -2x^2 \).

Solution to part 2:
\[ f(x) = 4\sum_{n=0}^{\infty}(-2x^2)^n = 4\sum_{n=0}^{\infty}(-2)^n x^{2n} = \sum_{n=0}^{\infty}(-1)^n \cdot 2^{n+2}\, x^{2n}. \]

Convergence requires \( |-2x^2| < 1 \), i.e., \( x^2 < \dfrac{1}{2} \), so \( |x| < \dfrac{1}{\sqrt{2}} \). The interval of convergence is \( \left(-\dfrac{1}{\sqrt{2}},\, \dfrac{1}{\sqrt{2}}\right) \).

Solution to part 1:
\[ \frac{1}{x} = \frac{1}{1-(-(x-1))} = \sum_{n=0}^{\infty}[-(x-1)]^n = \sum_{n=0}^{\infty}(-1)^n (x-1)^n. \]
Solution to part 2:

Convergence requires \( |-(x-1)| = |x-1| < 1 \), i.e., \( x \in (0, 2) \). At \( x = 0 \): \( \sum (-1)^n(-1)^n = \sum 1 \), which diverges. At \( x = 2 \): \( \sum (-1)^n \), which diverges. The interval of convergence is \( (0,\, 2) \).

Solution to part 1:

Set \( 3 = A(1+2x) + B(1-x) \). At \( x = 1 \): \( 3 = 3A \Rightarrow A = 1 \). At \( x = -\tfrac{1}{2} \): \( 3 = \tfrac{3}{2}B \Rightarrow B = 2 \).

\[ f(x) = \frac{1}{1-x} + \frac{2}{1+2x}. \]
Solution to part 2:
\[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \quad |x| < 1. \] \[ \frac{2}{1+2x} = \frac{2}{1-(-2x)} = 2\sum_{n=0}^{\infty}(-2x)^n = \sum_{n=0}^{\infty}(-1)^n 2^{n+1} x^n, \quad |x| < \tfrac{1}{2}. \]

Combining:

\[ f(x) = \sum_{n=0}^{\infty}\bigl[1 + (-1)^n 2^{n+1}\bigr]\,x^n. \]
Solution to part 3:

The first series converges for \( |x| < 1 \) and the second for \( |x| < \dfrac{1}{2} \). The combined series is valid on the intersection: \( \left(-\dfrac{1}{2},\, \dfrac{1}{2}\right) \).

Solution to part 1:
\[ \frac{d}{dx}\left(\sum_{n=0}^{\infty} x^n\right) = \sum_{n=1}^{\infty} n\,x^{n-1} = \frac{1}{(1-x)^2}, \quad |x| < 1. \] Solution to part 2:

Multiply through by \( x \):

\[ \frac{x}{(1-x)^2} = \sum_{n=1}^{\infty} n\,x^n = x + 2x^2 + 3x^3 + \cdots, \quad |x| < 1. \]

The interval of convergence is \( (-1,\,1) \) (differentiation preserves the radius; endpoints must be rechecked, and both yield divergent series here).

Solution to part 1:
\[ \ln(1+x) = \int_0^x \frac{1}{1+t}\,dt = \sum_{n=0}^{\infty}(-1)^n \int_0^x t^n\,dt = \sum_{n=0}^{\infty}(-1)^n \frac{x^{n+1}}{n+1} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\,x^n. \]

The radius of convergence is 1. At \( x = 1 \) the series is the alternating harmonic series, which converges. At \( x = -1 \) it is \( -\sum 1/n \), which diverges. Hence the interval of convergence is \( (-1,\,1] \).

Solution to part 2:

Set \( x = 1 \) (which is in the interval of convergence):

\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} = \ln(1+1) = \ln 2. \]

Solution to part 1:
\[ e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!} = 1 – x^2 + \frac{x^4}{2!} – \frac{x^6}{3!} + \cdots \]

This converges for all \( x \in \mathbb{R} \).

Solution to part 2:
\[ \int_0^{0.5} e^{-x^2}\,dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_0^{0.5} x^{2n}\,dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!\,(2n+1)}\left(\frac{1}{2}\right)^{2n+1}. \]
Solution to part 3:

The first four terms (\( n = 0, 1, 2, 3 \)):

Partial sum: \( 0.500000 – 0.041667 + 0.003125 – 0.000186 \approx 0.461272 \).

The next term (the error bound) is \( \dfrac{1}{24\cdot 9}\cdot\dfrac{1}{512} \approx 0.0000090 \). Thus:

\[ \int_0^{0.5} e^{-x^2}\,dx \approx 0.4613 \quad \text{with error} < 0.000009. \]

Solution to part 1:

The pattern repeats: \( f^{(2k)}(0) = (-1)^k \), and all odd-order terms vanish.

Solution to part 2:
\[ \cos x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\,x^{2n} = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + \cdots \]

Since the Ratio Test yields \( R = \infty \), the interval of convergence is \( (-\infty, +\infty) \).

Solution to part 1:

For \( n \geq 1 \): \( f^{(n)}(x) = \dfrac{(-1)^{n-1}(n-1)!}{x^n} \), so \( f^{(n)}(1) = (-1)^{n-1}(n-1)! \). Also \( f(1) = \ln 1 = 0 \).

Solution to part 2:
\[ \ln x = \sum_{n=1}^{\infty}\frac{f^{(n)}(1)}{n!}(x-1)^n = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n. \]

The Ratio Test gives \( R = 1 \), so the open interval is \( (0, 2) \). At \( x = 2 \): \( \displaystyle\sum \frac{(-1)^{n-1}}{n} = \ln 2 \), which converges. At \( x = 0 \): \( \displaystyle\sum \frac{-1}{n} \), which diverges. The interval of convergence is \( (0,\, 2] \).

Solution to part 1:
\[ P_5(x) = x – \frac{x^3}{3!} + \frac{x^5}{5!} = x – \frac{x^3}{6} + \frac{x^5}{120}. \]
\[ P_5(0.3) = 0.3 – \frac{(0.3)^3}{6} + \frac{(0.3)^5}{120} = 0.3 – \frac{0.027}{6} + \frac{0.00243}{120} \]
\[ = 0.3 – 0.004500 + 0.0000202 = 0.2955202. \]
Solution to part 2:

The remainder after the degree-5 polynomial involves \( f^{(6)}(x) = -\sin x \) or \( f^{(7)}(x) = -\cos x \). Since \( |f^{(6)}(x)| \leq 1 \) for all \( x \):

\[ |R_5(0.3)| \leq \frac{1}{6!}(0.3)^6 = \frac{(0.3)^6}{720} = \frac{0.000729}{720} \approx 1.0 \times 10^{-6}. \]
Solution to part 3:

The error bound is less than \( 10^{-6} \), confirming accuracy to at least six decimal places. (The true value \( \sin(0.3) \approx 0.295520207 \) agrees with \( P_5(0.3) \approx 0.295520 \) to six decimal places.)

Solution to part 1:
\[ 1 – \cos x – \frac{x^2}{2} = 1 – \left(1 – \frac{x^2}{2} + \frac{x^4}{24} – \frac{x^6}{720}+\cdots\right) – \frac{x^2}{2} \]
\[ = 1 – 1 + \frac{x^2}{2} – \frac{x^4}{24} + \cdots – \frac{x^2}{2} = -\frac{x^4}{24} + O(x^6). \]
Solution to part 2:
\[ \frac{1 – \cos x – \tfrac{x^2}{2}}{x^4} = \frac{-\tfrac{x^4}{24} + O(x^6)}{x^4} = -\frac{1}{24} + O(x^2) \to \boxed{-\frac{1}{24}} \quad \text{as } x\to 0. \]

Solution to part 1:

Substituting \( y = \displaystyle\sum_{n=0}^{\infty} a_n x^n \) and \( y’ = \displaystyle\sum_{n=1}^{\infty} n\,a_n x^{n-1} \):

\[ \sum_{n=1}^{\infty} n\,a_n x^{n-1} = 2x \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} 2a_n x^{n+1} = \sum_{n=1}^{\infty} 2a_{n-1}\,x^n. \]

Re-indexing the left side (let \( m = n-1 \)):

\[ \sum_{m=0}^{\infty}(m+1)a_{m+1}\,x^m = \sum_{m=1}^{\infty}2a_{m-1}\,x^m. \]

Comparing coefficients: \( a_1 = 0 \) (constant term on right is 0), and for \( m \geq 1 \): \( (m+1)a_{m+1} = 2a_{m-1} \).

Solution to part 2:

With \( a_0 = 1 \) and \( a_1 = 0 \), the recurrence gives:

\[ a_2 = \frac{2a_0}{2} = 1,\quad a_3 = \frac{2a_1}{3} = 0,\quad a_4 = \frac{2a_2}{4} = \frac{1}{2},\quad a_6 = \frac{2a_4}{6} = \frac{1}{6},\ldots \]

In general, \( a_{2k} = \dfrac{1}{k!} \) and all odd-indexed coefficients are zero. The first five non-zero terms:

\[ y = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \frac{x^8}{4!} + \cdots \]
Solution to part 3:

This is precisely the Maclaurin series for \( e^{x^2} \), since:

\[ e^{x^2} = \sum_{k=0}^{\infty}\frac{(x^2)^k}{k!} = \sum_{k=0}^{\infty}\frac{x^{2k}}{k!}. \]

The closed-form solution is \( \boxed{y = e^{x^2}} \), which can be verified directly: \( y’ = 2x e^{x^2} = 2xy \) and \( y(0) = 1 \). \(\checkmark\)