Series is a very practical branch of mathematics, and series summation is one of the core contents of series research. Many problems can be expressed as the sum or product of a series, namely operations of the type \sum_{i=1}^n f(i) or \prod_{i=1}^n f(i). Since \ln\left(\prod_{i=1}^n f(i)\right) = \sum_{i=1}^n \ln(f(i)) = k, it follows that \prod_{i=1}^n f(i) = e^k. In other words, the product of a series can also be calculated by converting it into a series summation. Therefore, we can focus our efforts on series summation.
To solve general series summation problems, we consider the solution to the following equation: f(x+\epsilon) - f(x) = g(x) \tag{1} where g(x) is a known function of x, \epsilon is a constant, and the initial condition is f(k)=b. We seek the expression for f(x).
Expanding f(x+\epsilon) using a Taylor series, we get: f(x+\epsilon) = f(x) + f'(x)\epsilon + \frac{1}{2} f''(x)\epsilon^2 + \frac{1}{6} f'''(x)\epsilon^3 + \dots Substituting this into the original equation, we obtain: f'(x)\epsilon + \frac{1}{2} f''(x)\epsilon^2 + \frac{1}{6} f'''(x)\epsilon^3 + \dots = g(x) \tag{2} Integrating term by term yields: f(x)\epsilon + \frac{1}{2} f'(x)\epsilon^2 + \frac{1}{6} f''(x)\epsilon^3 + \frac{1}{24} f'''(x)\epsilon^4 + \dots = \int g(x)dx Rearranging this, we get: f(x)\epsilon = \int g(x)dx - \left(\frac{1}{2} f'(x)\epsilon^2 + \frac{1}{6} f''(x)\epsilon^3 + \frac{1}{24} f'''(x)\epsilon^4 + \dots\right) \tag{3} Differentiating each term of (1), we get: f'(x+\epsilon) - f'(x) = g'(x) \tag{4} From formula (2), we can derive: f'(x)\epsilon = \int g'(x)dx - \left(\frac{1}{2} f''(x)\epsilon^2 + \frac{1}{6} f'''(x)\epsilon^3 + \frac{1}{24} f^{(4)}(x)\epsilon^4 + \dots\right) \tag{5} Substituting (5) into (3) yields: \begin{aligned} f(x)\epsilon &= \int g(x)dx - \frac{1}{2} \left[g(x) - \left(\frac{1}{2} f''(x)\epsilon^2 + \frac{1}{6} f'''(x)\epsilon^3 + \frac{1}{24} f^{(4)}(x)\epsilon^4 + \dots\right)\right]\epsilon^2 \\ &\quad + \frac{1}{6} f''(x)\epsilon^3 + \frac{1}{24} f'''(x)\epsilon^4 + \dots \\ &= \int g(x)dx - \frac{1}{2} g(x) + \sum_{i=2}^{\infty} \left(\frac{1}{2} \cdot \frac{1}{i!} \epsilon^{i+2} - \frac{1}{(i+1)!} \epsilon^{i+1}\right) f^{(i)}(x) \end{aligned} \tag{6}
The above derivation provides some inspiration: we can find that the solution to the equation f(x+\epsilon) - f(x) = g(x) can be expressed as the series: \epsilon f(x) = C + a_0 \int g(x)dx + a_1 g(x) + a_2 g'(x) + a_3 g''(x) \dots \tag{7} where C and a_n are constants that depend only on \epsilon. Substituting this solution into (2), we get: \begin{aligned} &a_0 g(x) + a_1 g'(x) + a_2 g''(x) + a_3 g'''(x) \dots + \\ &\frac{1}{2} \cdot (a_0 g'(x) + a_1 g''(x) + a_2 g'''(x) + a_3 g^{(4)}(x) \dots) \epsilon \\ &+ \frac{1}{6} \cdot (a_0 g''(x) + a_1 g'''(x) + a_2 g^{(4)}(x) + a_3 g^{(5)}(x) \dots) \epsilon^2 + \dots = g(x) \end{aligned} Rearranging terms: \begin{aligned} &a_0 g(x) + (a_1 + \frac{1}{2} a_0 \epsilon) g'(x) + (a_2 + \frac{1}{2} a_1 \epsilon + \frac{1}{6} a_0 \epsilon^2) g''(x) + \\ &(a_3 + \frac{1}{2} a_2 \epsilon + \frac{1}{6} a_1 \epsilon^2 + \frac{1}{24} a_0 \epsilon^3) g'''(x) + \dots = g(x) \end{aligned} According to the principle of series equality, we must have: a_0 = 1, \quad \sum_{i=0}^n \left(a_i \epsilon^{n-i} \cdot \frac{1}{(n+1-i)!}\right) = 0 Or written as: a_0 = 1, \quad a_n = -\sum_{i=0}^{n-1} \left(a_i \epsilon^{n-i} \cdot \frac{1}{(n+1-i)!}\right) \tag{8} At this point, we have found the general solution to the equation f(x+\epsilon) - f(x) = g(x). Based on the initial conditions, we can determine the unique solution: \epsilon f(x) = b\epsilon + \left[a_0 \int g(x)dx + a_1 g(x) + a_2 g'(x) + a_3 g''(x) \dots \right]_k^x
After this long discussion, some readers might feel confused: what does this have to do with series? There is indeed a connection. For example:
Let S(n) = \sum_{i=1}^n f(i), then S(n+1) - S(n) = f(n+1).
Now you understand: if the general term formula is known, then series summation is equivalent to the case where \epsilon=1 in equation (1)! For convenience in application, we list the coefficients for \epsilon=1 from equation (8):
a_0 = 1
a_1 = -1/2, a_2 = 1/12
a_3 = 0, a_4 = -1/720
a_5 = 0, a_6 = 1/30240
a_7 = 0, a_8 = -1/1209600
a_9 = 0, a_{10} = 1/47900160
a_{11} = 0, a_{12} = -691/1307674368000
......
As can be seen, the convergence is rapid. This summation method is the famous “Euler-Maclaurin summation formula.” Based on the results above, let us try to find some summation formulas, such as \sum_{x=1}^n x^3, following these steps: \begin{aligned} S(n+1) - S(n) &= (n+1)^3, \quad S(1) = 1 \\ \int (n+1)^3 dn &= \frac{1}{4} (n+1)^4 \\ \frac{d(n+1)^3}{dn} &= 3(n+1)^2 \\ \frac{d^2(n+1)^3}{dn^2} &= 6(n+1) \\ \frac{d^3(n+1)^3}{dn^3} &= 6 \end{aligned} Therefore: \begin{aligned} \sum_{x=1}^n x^3 &= S(n) \\ &= 1 + \left[\frac{1}{4} (n+1)^4 + (n+1)^3 \cdot (-1/2) + 3(n+1)^2 \cdot (1/12) + 6 \cdot (-1/720)\right]_1^n \\ &= \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4} \end{aligned}
Additionally, we can find an approximate formula for n!. Since \ln(n!) = \sum_{x=1}^n \ln x, we solve S(x+1) - S(x) = \ln(x+1). Since \int \ln(x+1) dx = (x+1) \ln(x+1) - (x+1) + \text{Const}, naturally, the first-order approximation is n! \approx e^{(x+1) \ln(x+1) - (x+1) + 2(1-\ln 2)}. We can continue to write: \begin{aligned} \sum_{x=1}^n \ln x &= C_0 + \left[(x+1) \ln(x+1) - (x+1) - \frac{1}{2} \ln(x+1) + \right. \\ &\quad \left. \frac{1}{12(x+1)} - \frac{1}{360(x+1)^3} + \frac{1}{1260(x+1)^5} - \frac{1}{1680(x+1)^7} \dots \right]_k^n \end{aligned} where C_0 = \sum_{x=1}^k \ln x. It is evident that this formula is convergent.
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