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Lemma: in any ring $R$, any power series in $R[[x]]$ with constant term $1$ is invertible. Explanation. Every power series with constant term $1$ can be written as $1-xf(x)$ for some other power series $f(x)$. Then $(1-xf(x))(1+xf(x)+x^2f(x)^2+\cdots)=1$. The geometric series can be simplified to a power series in $x$ by expanding all of the powers $f(x)^n$ ...

5

Pick any $B>0$. Then, if $|x|\leq B$, then one has that $|f_k(x)|=|x|^k/k!\leq B^k/k!$ for each non-negative integer $k$. Since $$\sum_{k=0}^{\infty}\frac{B^k}{k!}=\exp(B)$$ is convergent, Weierstrass's $M$-test reveals that the series $\sum_{k=0}^{\infty} x^k/k!$ converges uniformly for $x\in[-B,B]$. Together with the continuity of the partial sums ...

3

The number of solutions of the equation equals the coefficient of $z^n$ in the expression $$\frac{1}{(1-z)(1-z^2)(1-z^3)\ldots }=\sum_{n}p(n)z^n,$$ where $p(n)$ is the Euler partition function, so that the number of combinations is $p(n)$. (Christoph): There is a bijection between the partition function and the desired function, since from a partition ...

3

Almost certainly not: if there were a "general formula" for $\sum_{i=1}^r p_i^n$ for even a single $n$, you'd also have a "general formula" for the prime numbers by $$p_r=\sqrt[n]{\left(\text{“general formula” for }\sum_{i=1}^r p_i^n\right) - \left(\text{“general formula” for }\sum_{i=1}^{r-1}p_i^n\right)}$$

2

We have: \begin{align}&f(x)=\sum_{n=0}^\infty\binom\alpha n x^n\\{}\\ &f'(x)=\sum_{n=1}^\infty\binom\alpha n nx^{n-1}\end{align} Perhaps the trick you're looking for is $$n\binom\alpha n=\alpha\binom{\alpha-1}{n-1}$$ Added on request: $$\alpha f(x)=\sum_{n=0}^\infty \alpha\binom\alpha ... 1 Taylor expanding \cos z about z=\frac{\pi}{2} one finds that: $$\cos z=-\left(z-\frac{\pi}{2}\right)+\frac{1}{6}\left(z-\frac{\pi}{2}\right)^3+\ldots$$ Similarly, Taylor expanding \frac{1}{z+\frac{\pi}{2}} about z=\frac{\pi}{2} one finds that: ... 1 A very simple proof of this actually involves complex analysis. The set of complex numbers extends to the reals. e^{z}=\sum_{0}^ \infty \frac{z^{k}}{k!}  This is a series for e^{z} expanded at \alpha=0. In complex analysis in order for a number to be inside the disc of convergence it has to follow the following conditions: Let \beta be the ... 1 We will use the fact that$$ \sum_{n=0}^\infty\frac{x^n}{n!} $$converges absolutely for all x\in\mathbb{R}. This can easily be shown using the ratio test and means that for any x, there is an N so that$$ \sum_{n=N+1}^\infty\frac{|x|^n}{n!}\le\epsilon $$Find an N so that we have$$ ...

1

One may recall that, as $x \to 0$, \begin{align} e^x&=\sum_{n=0}^{\infty}\frac{x^n}{n!}, \quad x \in \mathbb{C}, \tag1\\\\ \frac{1}{1-x^2}&=\sum_{n=0}^{\infty}x^{2n}=\frac1{2}\sum_{n=0}^{\infty}(1+(-1)^n)x^{n}, \quad |x|<1, \tag2 \end{align} then using the Cauchy product we get $$... 1 One method of evaluating \sum_{n=0}^\infty(1+n)x^n can be like this, we take the generating function f = \sum_{n=0}^\infty x^n , then$$\sum_{n=0}^\infty (n+1)x^n = (xD + 1) f  \frac{x}{(1-x)^2} + \frac{1}{1-x} = \frac{1}{(1-x)^2}here D means differentiation w.r.t x

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