Previous in Relation and Functions Next in Relation and Functions
Question Number 33886 by math khazana by abdo last updated on 26/Apr/18
$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{3}}. \\ $$
Commented by math khazana by abdo last updated on 29/Apr/18
$${changement}\:{of}\:{indice}\:{n}={p}−\mathrm{1}\:{give} \\ $$$${S}=\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} }{\mathrm{2}\left({p}−\mathrm{1}\right)+\mathrm{3}}\:=\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} }{\mathrm{2}{p}+\mathrm{1}} \\ $$$$=−\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{2}{p}+\mathrm{1}}\:\:=\:−\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{2}{p}+\mathrm{1}}\:+\mathrm{1} \\ $$$${w}\left({x}\right)=\:\sum_{{p}={o}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{2}{p}+\mathrm{1}}{x}^{\mathrm{2}{p}+\mathrm{1}} \:\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$${w}^{'} \left({x}\right)\:=\sum_{{p}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{p}} \:{x}^{\mathrm{2}{p}} \:=\sum_{{p}=\mathrm{0}} ^{\infty} \left(−{x}^{\mathrm{2}} \right)^{{p}} \:=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{w}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\lambda\:\:{but}\:\lambda\:={w}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${w}\left({x}\right)\:=\:{arctanx} \\ $$$${S}=\:\mathrm{1}\:−{w}\left(\mathrm{1}\right)\:=\:\mathrm{1}−\frac{\pi}{\mathrm{4}}\: \\ $$$$\bigstar{S}=\mathrm{1}−\frac{\pi}{\mathrm{4}}\bigstar \\ $$