高専の数学解説

第3章 3.1

(1)

∑n=1∞xnn\sum_{n=1}^{\infty} \frac{x^n}{n}

r=lim⁡n→∞∣1n1n+1∣=lim⁡n→∞∣n+1n∣=lim⁡n→∞∣1+1n∣=1r = \lim_{n \to \infty} \left| \frac{\dfrac{1}{n}}{\dfrac{1}{n+1}} \right| = \lim_{n \to \infty} \left| \frac{n+1}{n} \right| = \lim_{n \to \infty} \left| 1 + \frac{1}{n} \right| = 1

(2)

∑n=0∞(−1)n(n+1)xn\sum_{n=0}^{\infty} (-1)^n (n+1) x^n

r=lim⁡n→∞∣(−1)n(n+1)(−1)n(n+2)∣=lim⁡n→∞∣1+1n1+2n∣=1r = \lim_{n \to \infty} \left| \frac{(-1)^n (n+1)}{(-1)^n (n+2)} \right| = \lim_{n \to \infty} \left| \frac{1 + \dfrac{1}{n}}{1 + \dfrac{2}{n}} \right| = 1

(3)

∑n=1∞xnn\sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n}}

r=lim⁡n→∞∣1n1n+1∣=lim⁡n→∞∣n+1n∣=1r = \lim_{n \to \infty} \left| \frac{\dfrac{1}{\sqrt{n}}}{\dfrac{1}{\sqrt{n+1}}} \right| = \lim_{n \to \infty} \left| \frac{\sqrt{n+1}}{\sqrt{n}} \right| = 1

(4)

∑n=0∞xn2n\sum_{n=0}^{\infty} \frac{x^n}{2^n}

r=lim⁡n→∞∣12n12n+1∣=lim⁡n→∞∣2n+12n∣=2r = \lim_{n \to \infty} \left| \frac{\dfrac{1}{2^n}}{\dfrac{1}{2^{n+1}}} \right| = \lim_{n \to \infty} \left| \frac{2^{n+1}}{2^n} \right| = 2

(5)

∑n=1∞xn3nn2\sum_{n=1}^{\infty} \frac{x^n}{3^n n^2}

r=lim⁡n→∞∣13nn213n+1(n+1)2∣=lim⁡n→∞∣3n+1(n+1)23nn2∣=3r = \lim_{n \to \infty} \left| \dfrac{\dfrac{1}{3^n n^2}}{\dfrac{1}{3^{n+1}(n+1)^2}} \right| = \lim_{n \to \infty} \left| \frac{3^{n+1}(n+1)^2}{3^n n^2} \right| = 3

(6)

∑n=1∞xnnn\sum_{n=1}^{\infty} \frac{x^n}{n^n}

r=lim⁡n→∞∣1nn1(n+1)n+1∣=lim⁡n→∞∣(n+1)n+1nn∣分母分子に 1nn+1 を掛けるとr = \lim_{n \to \infty} \left| \dfrac{\dfrac{1}{n^n}}{\dfrac{1}{(n+1)^{n+1}}} \right| = \lim_{n \to \infty} \left| \frac{(n+1)^{n+1}}{n^n} \right| \qquad \text{分母分子に } \dfrac{1}{n^{n+1}} \text{ を掛けると}

=lim⁡n→∞∣(1+1n)n+11n∣=lim⁡n→∞∣n(1+1n)n+1∣=∞= \lim_{n \to \infty} \left| \dfrac{\left(1+\dfrac{1}{n}\right)^{n+1}}{\dfrac{1}{n}} \right| = \lim_{n \to \infty} \left| n\left(1+\frac{1}{n}\right)^{n+1} \right| = \infty

解説: r31bn1z