Question Number 66324 by mathmax by abdo last updated on 12/Aug/19
$${find}\:{nature}\:{of}\:{the}\:{serie}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}^{{n}} \:+{ln}\left({n}\right)} \\ $$
Commented by mathmax by abdo last updated on 21/Aug/19
$${we}\:{have}\:{S}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {u}_{{n}} \:\:{with}\:{u}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} \:+{ln}\left({n}\right)} \\ $$$${we}\:{have}\:{u}_{{n}} >\mathrm{0}\:\:\:{and}\:{u}_{{n}} \rightarrow\mathrm{0}\:\:{let}\:{prove}\:{that}\:\left({u}_{{n}} \right){is}\:{decreasing} \\ $$$${let}\:{f}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{t}} \:+{ln}\left({t}\right)}\:\:{with}\:{t}\geqslant\mathrm{1}\:\:{we}\:{have}\: \\ $$$${f}^{'} \left({t}\right)\:=−\frac{\left(\mathrm{2}^{{t}} \:+{ln}\left({t}\right)\right)^{'} }{\left(\mathrm{2}^{{t}} \:+{lnt}\right)^{\mathrm{2}} }\:\:\:\:{and}\:\:\left(\mathrm{2}^{{t}} \:+{lnt}\right)^{'} \:=\left({e}^{{tln}\mathrm{2}} +{lnt}\right)^{'} ={ln}\left(\mathrm{2}\right)\mathrm{2}^{{t}} \:+\frac{\mathrm{1}}{{t}} \\ $$$$\Rightarrow{f}^{'} \left({t}\right)\:=−\frac{{ln}\left(\mathrm{2}\right)\mathrm{2}^{{t}} \:+\frac{\mathrm{1}}{{t}}}{\left(\mathrm{2}^{{t}} \:+{lnt}\right)^{\mathrm{2}} }<\mathrm{0}\:\Rightarrow{f}\:{decrease}\:{on}\:\left[\mathrm{1},+\infty\left[\:\:{so}\:{S}_{{n}} {is}\right.\right. \\ $$$${a}\:{alternate}\:{serie}\:\Rightarrow\:{S}_{{n}} \:{converges}. \\ $$