\sum_{i=0}^{\infty} a_i x^i=a_0+a_1 x+a_2 x^2+a_3 x^3+\dots
Recently, for math competitions, I have been studying problems related to sequences and combinatorics. Since I dislike designing specific tricks for specific problems, I prefer general methods, even if the process is relatively cumbersome. Therefore, I have a special fondness for mathematical induction (recursion method) and the generating function method. The former only requires listing the recursive relationship of the problem without detailed analysis, ultimately transforming the problem into solving functional equations. The latter skillfully creates a one-to-one correspondence between the sequence \{a_n\} and the power series \sum_{i=0}^{\infty} a_i x^i, obtaining results through algebraic or calculus operations. Here, we do not need to consider the convergence or divergence of the series; we only need to know which “generating function” it corresponds to (the generating function yields the series \sum_{i=0}^{\infty} a_i x^i after Taylor expansion). Clearly, both methods ultimately reduce the problem to an algebraic one.
The generating function method was first applied to probability by Laplace and was later generalized, becoming a widely used method. Because it uses algebraic methods, it is highly extensible. In the future, I will spend some time discussing the applications of the generating function method with everyone. However, here we will first study a problem: the generating function method generally corresponds to an infinite series, so given the general term formula for a_n, we need to find the function corresponding to \sum_{i=0}^{\infty} a_i x^i. Below, we will study \sum_{n=1}^\infty n^m x^n, m \in \mathbb{N}_+, where m is a constant. That is, 1^m x + 2^m x^2 + 3^m x^3 + \dots.
From calculus, we have already obtained: \sum_{n=1}^\infty n^0 x^n = x + x^2 + x^3 + \dots = \frac{1}{1-x} - 1
And we have: \sum_{n=1}^\infty n^m x^n = x \sum_{n=1}^\infty n^m x^{n-1} = x \sum_{n=1}^\infty \frac{d(n^{m-1} x^n)}{dx} = x \frac{d}{dx} \sum_{n=1}^\infty n^{m-1} x^n
Through this transformation, we have “reduced the degree” of n^m to n^{m-1}. Following this idea, we can obtain a recurrence formula. Let T_m = \sum_{n=1}^\infty n^m x^n, then:
T_1 = \frac{x}{1-x} T_{m+1} = x \frac{dT_m}{dx}
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