一道简单的无穷级数题目
- 求级数 ∑ n = 1 + ∞ n x n \sum _{n=1} ^ {+\infty} n x^n n=1∑+∞nxn
解析：
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s = \sum _{n=1} ^ {+\infty} n x^n
s=n=1∑+∞nxn
s 1 = ∑ n = 1 + ∞ n x n − 1 s_1 = \sum _{n=1} ^ {+\infty} n x^{n-1} s1=n=1∑+∞nxn−1
则 s = s 1 x s = s_1 x s=s1x
∫ s 1 d x = ∑ n = 1 n = ∞ x n = 1 1 − x − 1 = x 1 − x \int s_1 dx = \sum_{n=1}^{n = \infty}x^n = \frac{1}{1-x} - 1 = \frac{x}{1-x} ∫s1dx=n=1∑n=∞xn=1−x1−1=1−xx
且x的收敛域为(0,1)
s 1 = ( x 1 − x ) ′ = 1 ( 1 − x ) 2 s_1 = (\frac{x}{1-x})' = \frac{1}{(1-x)^2} s1=(1−xx)′=(1−x)21
s = s 1 x = x ( 1 − x ) 2 x ∈ ( 0 , 1 ) s = s_1 x =\frac{x}{(1-x)^2} \quad x \in (0,1) s=s1x=(1−x)2xx∈(0,1)
解出上题中需要知道的一个重要的知识点是 1 1 − x \frac{1}{1-x} 1−x1的泰勒展开式：
1 1 − x = 1 + 1 1 ! ( 1 ( 1 − x ) 2 ∣ x = 0 ) x + 1 2 ! ( 2 ( 1 − x ) 3 ∣ x = 0 ) x 2 + + 1 3 ! ( 3 ! ( 1 − x ) 4 ∣ x = 0 ) x 3 + 1 4 ! ( 4 ! ( 1 − x ) 5 ∣ x = 0 ) x 4 + 1 5 ! ( 5 ! ( 1 − x ) 6 ∣ x = 0 ) x 5 + . . . + 1 ( n − 1 ) ! ( ( n − 1 ) ! ( 1 − x ) n ∣ x = 0 ) x n − 1 + 1 n ! ( n ! ( 1 − x ) n + 1 ∣ x = 0 ) x n = 1 + x + x 2 + x 3 + x 4 + x 5 + . . . + x n − 1 + x n = ∑ n = 0 + ∞ x n \frac{1}{1-x} = 1 + \frac{1}{1!} (\frac{1}{(1-x)^2}| _{x = 0}) x+\frac{1}{2!} (\frac{2}{(1-x)^3}| _{x = 0})x^2 + \\ +\frac{1}{3!}(\frac{3!}{(1-x)^4}| _{x = 0})x^3+\frac{1}{4!}(\frac{4!}{(1-x)^5}| _{x = 0})x^4+\frac{1}{5!}(\frac{5!}{(1-x)^6}| _{x = 0})x^5+...\\ +\frac{1}{(n-1)!}(\frac{(n-1)!}{(1-x)^n}| _{x = 0})x^{n-1}+\frac{1}{n!}(\frac{n!}{(1-x)^{n+1}}| _{x = 0})x^{n} = \\ 1 + x + x^2 + x^3 + x^4 + x^5 + ... + x^{n-1} + x^n = \sum_{n=0}^{+\infty} x^n 1−x1=1+1!1((1−x)21∣x=0)x+2!1((1−x)32∣x=0)x2++3!1((1−x)43!∣x=0)x3+4!1((1−x)54!∣x=0)x4+5!1((1−x)65!∣x=0)x5+...+(n−1)!1((1−x)n(n−1)!∣x=0)xn−1+n!1((1−x)n+1n!∣x=0)xn=1+x+x2+x3+x4+x5+...+xn−1+xn=n=0∑+∞xn
上述式子是等比数列，根据等比数列的性质，公比x要小于1才收敛。当然根绝达朗贝尔判别法，也可以得出收敛半径是(0,1)
- 求极限 lim x → 0 sin ( sin x ) − x x 3 \lim_{x \to 0} \frac{\sin(\sin x) - x}{x^3} x→0limx3sin(sinx)−x
解：
lim x → 0 sin ( sin x ) − x x 3 = lim x → 0 sin x − sin 3 x 6 − x x 3 = lim x → 0 x − x 3 6 − x 3 6 − x x 3 = − 1 / 3 \lim_{x \to 0} \frac{\sin(\sin x) - x}{x^3} =\lim_{x \to 0} \frac{\sin x - \frac{\sin ^ 3x}{6}- x}{x^3} = \\ \lim_{x \to 0} \frac{x - \frac{x^3}{6} - \frac{x^3 }{6}- x}{x^3} = -1/3 x→0limx3sin(sinx)−x=x→0limx3sinx−6sin3x−x=x→0limx3x−6x3−6x3−x=−1/3
注意：在计算 sin 3 x 6 \frac{\sin ^ 3x}{6} 6sin3x的泰勒展开时要小心，不要丢项。