10.7: Power Series

Let \(a_n = (x - 2)^n/n.\) The key idea is to write a condition that assumes \(\sum_{n = 1}^\infty a_n\) is convergent, and then solve for \(x\) that satisfies the condition. We do so using the Ratio Test (see Section 10.6), which states that \(\sum_{n = 1}^\infty a_n\) converges if \(\lim_{n \to \infty} \abs{a_{n + 1}/a_n} \lt 1.\) The values of \(x\) that satisfy this inequality make up the interval of convergence.

Radius of Convergence Note that \(a_{n + 1} = (x - 2)^{n + 1}/(n + 1).\) We have \[ \ba L &= \lim_{n \to \infty} \abs{\frac{a_{n + 1}}{a_n}} \nl &= \lim_{n \to \infty} \abs{\dfrac{\dfrac{(x - 2)^{n + 1 \strut}}{n + 1}}{\dfrac{(x - 2)^n \strut}{n}}} \nl &= \lim_{n \to \infty} \abs{\frac{\cancel{(x - 2)^n} (x - 2)}{n + 1} \cdot \frac{n}{\cancel{(x - 2)^n}}} \nl &= \lim_{n \to \infty} \abs{\frac{(x - 2) \, n}{n + 1}} \nl &= \abs{x - 2} \lim_{n \to \infty} \frac{n}{n + 1} \nl &= \abs{x - 2} \pd \ea \] The series \(\sum_{n = 1}^\infty a_n\) converges if \(L \lt 1,\) that is, if \(\abs{x - 2} \lt 1.\) Solving this inequality shows \[-1 \lt x - 2 \lt 1 \iffArrow 1 \lt x \lt 3 \pd\] The radius of convergence is \(\boxed{R = 1}\) because the endpoints of \((1, 3)\) extend \(1\) in either direction of the center of the power series, \(2.\)

Interval of Convergence The preliminary interval of convergence is \(1 \lt x \lt 3.\) We now test each endpoint: Substituting \(x = 1\) into \(\sum_{n = 1}^\infty a_n\) gives \[\sum_{n = 1}^\infty \frac{(1 - 2)^n}{n} = \sum_{n = 1}^\infty \frac{(-1)^n}{n} \cma \] which converges by the Alternating Series Test because \(1/n\) is decreasing to \(0\) as \(n \to \infty.\) But at \(x = 3\) the series becomes \[\sum_{n = 1}^\infty \frac{(3 - 2)^n}{n} = \sum_{n = 1}^\infty \frac{(1)^n}{n} = \sum_{n = 1}^\infty \frac{1}{n} \cma \] which diverges because it is the Harmonic series. Therefore, we include \(x = 1\) and exclude \(x = 3\) in the interval of convergence, attaining \[\boxed{1 \leq x \lt 3}\]

Let \(a_n = x^{2n}/n! \,.\) Note that \(a_{n + 1} = x^{2n + 2}/(n + 1)! \, .\) We see \[ \ba L &= \lim_{n \to \infty} \abs{\frac{a_{n + 1}}{a_n}} \nl &= \lim_{n \to \infty} \abs{\frac{x^{2n + 2}}{(n + 1)!} \cdot \frac{n!}{x^{2n}}} \nl &= \lim_{n \to \infty} \abs{\frac{x^{2n} \, x^2}{(n + 1) \, n!} \cdot \frac{n!}{x^{2n}}} \nl &= \lim_{n \to \infty} \abs{\frac{x^2}{n + 1}} \nl &= \abs{x^2} \lim_{n \to \infty} \frac{1}{n + 1} \nl &= 0 \pd \ea \] Thus, \(L = 0 \lt 1\) is always true regardless of the value of \(x.\) Therefore, the power series \(\sum_{n = 1}^\infty a_n\) converges for any value of \(x.\) So the interval of convergence is \[\boxed{-\infty \lt x \lt \infty}\] and the radius of convergence is \(\boxed{R = \infty}.\)

We call \(a_n = (-1)^n \, x^{3n}/n.\) The series \(\sum_{n = 1}^\infty a_n\) converges if \(\lim_{n \to \infty} \abs{a_{n + 1}/a_n} \lt 1,\) according to the Ratio Test. We then solve for \(x\) that satisfies this condition. Note that \[a_{n + 1} = \frac{(-1)^{n + 1} \, x^{3(n + 1)}}{n + 1} = \frac{(-1)^{n + 1} \, x^{3n} \cdot x^3}{n + 1} \pd\]

Radius of Convergence We set up the limit \[ \ba L &= \lim_{n \to \infty} \abs{\frac{a_{n + 1}}{a_n}} \nl &= \lim_{n \to \infty} \abs{\frac{(-1)^{n + 1} \, x^{3n} \cdot x^3}{n + 1} \cdot \frac{n}{(-1)^n \, x^{3n}}} \nl &= \abs{x^3} \lim_{n \to \infty} \abs{\frac{n}{n + 1}} \nl &= \abs{x^3} \pd \ea \] The series converges if \(L \lt 1\)—that is, for \[\abs{x^3} \lt 1 \iffArrow \abs{x} \lt 1 \iffArrow -1 \lt x \lt 1 \pd\] This is our preliminary interval of convergence of \(\sum_{n = 1}^\infty a_n.\) The radius of convergence is \(\boxed{R = 1}.\)

Interval of Convergence We next evaluate \(\sum_{n = 1}^\infty a_n\) at the endpoints of \(-1 \lt x \lt 1.\) Substituting \(x = -1\) gives \[\sum_{n = 1}^\infty \frac{(-1)^n (-1)^{3n}}{n} = \sum_{n = 1}^\infty \frac{(-1)^{4n}}{n} = \sum_{n = 1}^\infty \frac{1}{n} \cma\] which diverges because it is the Harmonic series. At \(x = 1\) the series becomes \[\sum_{n = 1}^\infty \frac{(-1)^n (1)^{3n}}{n} = \sum_{n = 1}^\infty \frac{(-1)^n}{n} \cma\] which converges by the Alternating Series Test because \(\lim_{n \to \infty} (1/n) = 0\) and \(1/n\) is decreasing for \(n \geq 1.\) Thus, we exclude \(x = -1\) and include \(x = 1,\) so the interval of convergence is \[\boxed{-1 \lt x \leq 1}\]

Interval and Radius of Convergence The series converges if \(|r| \lt 1\)—that is, \[ \ba \abs{-x^2} \lt 1 &\iffArrow \abs{x^2} \lt 1 \nl &\iffArrow \abs{x} \lt 1 \nl &\iffArrow -1 \lt x \lt 1 \pd \ea \] Hence, the interval of convergence is \[\boxed{-1 \lt x \lt 1}\] The radius of convergence is \(\boxed{R = 1}.\)

We require a power series to be centered at \(x = 2,\) so the terms should contain the factor \((x - 2).\) In the denominator of \(g,\) we subtract \(2\) and add \(2\) to \(x \col\) \[g(x) = \frac{1}{1 - (x - 2 + 2)} = \frac{1}{-1 - (x - 2)} = \frac{-1}{1 - [-(x - 2)]} \pd\] This expression is attained by replacing \(x\) with \(-(x - 2)\) in \(\eqref{eq:geo-power-series},\) so \(g(x)\) represents an infinite geometric series with common ratio \(r = -(x - 2).\) Consequently, \[ \ba \frac{-1}{1 - [-(x - 2)]} &= - \sum_{n = 0}^\infty [-(x - 2)]^n \nl &= -\sum_{n = 0}^\infty (-1)^n (x- 2)^n \nl &= \boxed{\sum_{n = 0}^\infty (-1)^{n + 1} (x - 2)^n} \ea \]

Radius and Interval of Convergence The series converges when \(\abs r \lt 1,\) where \(r = -(x - 2).\) Thus, \(\abs r = \abs{x - 2},\) so we have \[\abs{x - 2} \lt 1 \iffArrow -1 \lt x - 2 \lt 1 \iffArrow 1 \lt x \lt 3 \pd\] Hence, the interval of convergence is \[\boxed{1 \lt x \lt 3}\] and the radius of convergence is \(\boxed{R = 1}.\)

1. Because the derivative of a sum is the sum of derivatives, we have \[f'(x) = \deriv{}{x} \sum_{n = 1}^\infty \frac{(x - 5)^n}{n} = \sum_{n = 1}^\infty \deriv{}{x} \parbr{\frac{(x - 5)^n}{n}} \pd\] We differentiate the summation formula, obtaining \[f'(x) = \boxed{\sum_{n = 1}^\infty (x - 5)^{n - 1}}\] We don't need to increment the starting index \(n = 1,\) because the first term of \(f\) is \((x - 5),\) \[f'(x) = \sum_{n = 0}^\infty (x - 5)^n \pd\]

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2. Since \(f(x) = g'(x),\) we have \(g(x) = \int f(x) \di x.\) Accordingly, we attain \(g\) by integrating the power series for \(f \col\) \[ \ba g(x) &= \sum_{n = 1}^\infty \int \frac{(x - 5)^n}{n} \di x \nl &= C + \sum_{n = 1}^\infty \frac{(x - 5)^{n + 1}}{n(n + 1)} \nl &= C + \frac{(x - 5)^2}{1(2)} + \frac{(x - 5)^3}{2(3)} + \frac{(x - 5)^4}{3(4)} + \cdots \pd \ea \] To solve for \(C,\) we substitute \(g(5) = 1\) and note that all the terms except \(C\) are \(0 \col\) \[ \ba g(5) = C + \frac{(5 - 5)^2}{1(2)} + \frac{(5 - 5)^3}{2(3)} + \cdots &= 1 \nl \implies C &= 1 \pd \ea \] Hence, \[g(x) = \boxed{1 + \sum_{n = 1}^\infty \frac{(x - 5)^{n + 1}}{n(n + 1)}}\]

Interval of Convergence; Solving for \(C\) The power series for \(1/(1 + x)\) converges if \(\abs{-x} \lt 1,\) so the radius of convergence is \(R = 1.\) Therefore, the power series for \(f(x) = \ln(1 + x),\) as in \(\eqref{eq:ln(1+x)},\) has a preliminary interval of convergence of \(-1 \lt x \lt 1.\) We exclude \(x = -1\) because \(f(-1) = \ln 0\) is undefined. But at \(x = 1\) \(\eqref{eq:ln(1+x)}\) becomes \[\sum_{n = 0}^\infty (-1)^n \frac{(1)^{n + 1}}{n + 1} = \sum_{n = 0}^\infty \frac{(-1)^n}{n + 1} \cma\] which converges by the Alternating Series Test since \(1/(n + 1)\) is decreasing to \(0.\) Thus, the interval of convergence is \[\boxed{-1 \lt x \leq 1}\] To solve for \(C\) in \(\eqref{eq:ln(1+x)},\) we substitute any value of \(x\) within the interval of convergence—say, \(x = 0.\) We then see \[ \ba \ln(1 + 0) &= C + 0 - \frac{0^2}{2} + \frac{0^3}{3} - \frac{0^4}{4} + \cdots \nl \implies C &= \ln 1 = 0 \pd \ea \] Thus, \(\eqref{eq:ln(1+x)}\) becomes \[\ln(1 + x) = \boxed{\sum_{n = 0}^\infty (-1)^n \frac{x^{n + 1}}{n + 1}}\]