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Common Summation and Series Formulas
During an exam a while ago, my mind went completely blank. I was supposed to find the sum of a series, but without even thinking, I tried to calculate it using an integral instead… I should review and organize the common formulas for series now.
1. polynomial
\[\sum_{k=1}^{n} k = \frac{1}{2}n(n+1)\]
\[\sum_{k=1}^{n} k^2 = \frac{1}{6}n(n+1)(2n+1)\]
\[\sum_{k=1}^{n} k^3 = \frac{1}{4}n^2(n+1)^2\]
\[\sum_{k=1}^{n} k^p = \frac{n^{p+1}}{p+1} + \frac{1}{2}n^p + \sum_{k=2}^{p} \frac{B_k}{k!}\, p^{\underline{k-1}}\, n^{p-k+1} \quad \text{where } p^{\underline{k-1}} = (p)_{k-1} = \frac{p!}{(p-k+1)!}\]
\[\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}\]
\[\sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90}\]
\[\sum_{k=1}^{\infty} \frac{1}{k^6} = \frac{\pi^6}{945}\]
\[\sum_{k=1}^{\infty} \frac{1}{k^{2n}} = (-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}\]
2. geometric
\[\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}, \quad \text{where } |x| < 1\]
\[\sum_{k=0}^{n} x^k = \frac{x^{n+1}-1}{x-1}, \quad \text{where } x \neq 1\]
\[\sum_{k=m}^{n} z^k = \frac{z^m - z^{n+1}}{1-z}, \quad z \neq 1\]
\[\sum_{k=1}^{n} kz^k = z\,\frac{1-(n+1)z^n + nz^{n+1}}{(1-z)^2}, \quad z \neq 1\]
\[\sum_{k=1}^{\infty} \frac{z^k}{k} = -\ln(1-z), \quad |z| < 1\]
\[\sum_{k=1}^{\infty} z^k = \frac{z}{1-z}, \quad |z| < 1\]
\[\sum_{k=1}^{\infty} kz^k = \frac{z}{(1-z)^2}, \quad |z| < 1\]
3. Harmonic
\[\sum_{n=1}^{k} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots > \ln(1+k) \quad \text{（看清楚符号）}\]
\[\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2\]
\[\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k-1} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \frac{\pi}{4}\]
4. others
\[\sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots = e\]
\[\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} = \frac{1}{1!} - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \frac{1}{9!} + \cdots = \sin 1\]
\[\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} = \frac{1}{0!} - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \frac{1}{8!} + \cdots = \cos 1\]
\[3 + \frac{4}{2\times3\times4} - \frac{4}{4\times5\times6} + \frac{4}{6\times7\times8} - \frac{4}{8\times9\times10} + \cdots = \pi\]
\[\sum_{k=1}^{\infty} \frac{1}{T_k} = \frac{1}{1} + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \cdots = 2\]
上面那个分母的规律是 $T_k = k + (k-1) + (k-2) + \cdots + 1$（三角形数）
\[\sum_{k=0}^{\infty} \frac{1}{(2k+1)(2k+2)} = \frac{1}{1\times2} + \frac{1}{3\times4} + \frac{1}{5\times6} + \frac{1}{7\times8} + \frac{1}{9\times10} + \cdots = \ln 2\]
\[\sum_{k=1}^{\infty} \frac{1}{2^k k} = \frac{1}{2} + \frac{1}{8} + \frac{1}{24} + \frac{1}{64} + \frac{1}{160} + \cdots = \ln 2\]
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