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Question Number 57232 by maxmathsup by imad last updated on 31/Mar/19

$$decomposetbefractionF\left(x\right)=\frac{1}{x^n(x+1)}withnintegrnatural.$$

Commented by maxmathsup by imad last updated on 14/Apr/19

$$thedcompositionofFisF\left(x\right)=\frac{a}{x+1}+\sum _{i=1}^n\frac{{\lambda }_i}{x^i}letw\left(x\right)=\frac{1}{x+1}$$$$letfind{D}_{n-1}\left(o\right)\left(w\right)\Rightarrow w\left(x\right)=\sum _{k=0}^{n-1}\frac{x^k}{k!}{w}^{\left(k\right)}\left(0\right)+\frac{x^n}{n!}\xi \left(x\right)(\xi \left(x\right)\to 0fx\to 0)$$$$but{w}^{\left(k\right)}\left(x\right)=\frac{{(-1)}^{k}k!}{{(x+1)}^{k+1}}\Rightarrow w^{\left(k\right)}\left(0\right)={(-1)}^{k}k!\Rightarrow w\left(x\right)=\sum _{k=0}^{n-1}{(-1)}^k{x}^k+\frac{x^n}{n!}\xi \left(x\right)$$$$\Rightarrow F\left(x\right)=\sum _{k=0}^{n-1}\frac{{(-1)}^k}{x^{n-k}}+\frac{1}{n!}\xi \left(x\right)changementofindicen-k=igive$$$$F\left(x\right)=\sum _{i=1}^n\frac{{(-1)}^{n-i}}{x^i}+\frac{1}{n!}\xi \left(x\right)butF\left(x\right)=\frac{a}{x+1}+\sum _{i=1}^n\frac{{\lambda }_i}{x^i}\Rightarrow {\lambda }_i={(-1)}^{n-i}$$$$i\in \left[[1,n]\right]wehavea={lim}_{x\to -1}(x+1)F\left(x\right)={(-1)}^n\Rightarrow$$$$F\left(x\right)=\frac{{(-1)}^n}{x+1}+\sum _{i=1}^n\frac{{(-1)}^{n-i}}{x^i}.$$$$$$

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