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Question Number 52679 by maxmathsup by imad last updated on 11/Jan/19

$$let{f}_n\left(x\right)=\frac{{(-1)}^n}{n+x}withx>0$$ $$1)studythesimpleconvergenceof\mathrm{\Sigma }{f}_n\left(x\right)$$ $$2)calculate{f}^{{}^{\prime }}\left(x\right)$$

Commented bymaxmathsup by imad last updated on 28/Feb/19

$$1)wehave\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{U}_n\left(x\right)with{U}_n\left(x\right)=\frac{1}{n+x}$$ $$forx>0fixed{U}_{n}isdecreasngand{lim}_{n\to +\mathrm{\infty }}{U}_n\left(x\right)=0so\mathrm{\Sigma }{f}_n\left(x\right)is$$ $$aalternateseriesothesimpleconvegenceisassured.$$ $$2)we\mid \mathrm{\Sigma }\frac{{(-1)}^n}{{(n+x)}^2}\mid ⩽\mathrm{\Sigma }\frac{1}{n^2}and\mathrm{\Sigma }{f}_n^{{}^{\prime }}\left(x\right)convergesunif.soifisitssum$$ $$f^{{}^{\prime }}\left(x\right)={\left(\mathrm{\Sigma }{f}_n\left(x\right)\right)}^{{}^{\prime }}=\mathrm{\Sigma }{f}_n^{{}^{\prime }}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^{n+1}}{{(n+x)}^2}$$

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