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Recitation 23

Example 1: Find a power series representation for the function f(x)=x2x2+1f(x) = \frac{x}{2x^2+1} and determine the interval of convergence.

Solution: Let y=−2x2y = -2x^2 in 1/(1−y)=∑n=0∞yn1/(1-y)=\sum_{n=0}^\infty y^n. We obtain 1/(1+2x2)=∑n=0∞(−2x2)n=∑n=0∞(−2)nx2n1/(1+2x^2)=\sum_{n=0}^\infty (-2x^2)^n = \sum_{n=0}^\infty (-2)^nx^{2n}, and then x/(2x2+1)=∑n=0∞(−2)nx2n+1x/(2x^2+1)=\sum_{n=0}^\infty (-2)^nx^{2n+1}. Because this is a geometric series, it converges when ∣−2x2∣<1|-2x^2| < 1, that is, ∣x∣<1/2|x| < 1/\sqrt{2}. Therefore the interval of convergence is (−1/2,1/2)(-1 / \sqrt{2}, 1/\sqrt{2}).

Example 2: Find a power series representation and the radius and the interval of convergence for (a) f(x)=ln⁡(x2+4)f(x) = \ln(x^2+4); (b) f(x)=tan⁡−1(2x)f(x)=\tan^{-1}(2x).

Hint: (a) Write out the power series of f′(x)=2x/(x2+4)f'(x) = 2x/(x^2+4) and integrate both sides; (b) Write out the power series of f′(x)=2/(1+4x2)f'(x) = 2/(1+4x^2) and integrate both sides.

Example 3: Find the radius of convergence and interval of convergence of the series ∑n=1∞n!xn1⋅3⋅5⋅…⋅(2n−1)\sum_{n=1}^\infty \frac{n!x^n}{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}.

Solution: Let an=n!xn1⋅3⋅5⋅…⋅(2n−1)a_n = \frac{n!x^n}{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}. Note that an+1an=(n+1)x2n+1\frac{a_{n+1}}{a_n} = \frac{(n+1)x}{2n+1} and it goes to x/2x/2 as n→∞n\to\infty. By the ratio test, when ∣x∣<2|x| < 2, the series converges. If ∣x∣=2|x| = 2, ∣an∣=2⋅4⋅6⋅…⋅(2n)/1⋅3⋅5⋅…⋅(2n−1)>1|a_n| = 2\cdot 4\cdot 6\cdot \ldots \cdot (2n) / 1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1) > 1. The limit test says the series diverges. Therefore the radius of convergence is 22 and the interval of convergence is (−2,2)(-2, 2).

Example 4: Test the series for convergence or divergence. (a) ∑n=2∞1nln⁡n\sum_{n=2}^\infty \frac{1}{n\sqrt{\ln n}}; (b) ∑n=2∞(−1)n−1n−1\sum_{n=2}^\infty\frac{(-1)^{n-1}}{\sqrt{n}-1}; (c) ∑n=1∞(nn+1)n2\sum_{n=1}^\infty\left(\frac{n}{n+1}\right)^{n^2}.

Hint: (a) Use the integral test; (b) Use the alternating series test; (c) Note that (nn+1)n2n=(nn+1)n=(1+1/n)−n=e−nln⁡(1+1/n)\sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}} = \left(\frac{n}{n+1}\right)^{n} = \left(1+1/n\right)^{-n} = e^{-n\ln(1+1/n)}. By l’Hospital’s rule, lim⁡x→∞xln⁡(1+1/x)=1\lim_{x\to\infty} x\ln(1+1/x) = 1, and so (nn+1)n2n→e−1<1\sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}}\to e^{-1} < 1 as n→∞n\to\infty. Use the root test.