Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 66324 by mathmax by abdo last updated on 12/Aug/19

find nature of the serie Σ_(n=1) ^∞   (((−1)^(n+1) )/(2^n  +ln(n)))

$${find}\:{nature}\:{of}\:{the}\:{serie}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}^{{n}} \:+{ln}\left({n}\right)} \\ $$

Commented by mathmax by abdo last updated on 21/Aug/19

we have S =−Σ_(n=0) ^∞ (−1)^n u_n   with u_n =(1/(2^n  +ln(n)))  we have u_n >0   and u_n →0  let prove that (u_n )is decreasing  let f(t) =(1/(2^t  +ln(t)))  with t≥1  we have   f^′ (t) =−(((2^t  +ln(t))^′ )/((2^t  +lnt)^2 ))    and  (2^t  +lnt)^′  =(e^(tln2) +lnt)^′ =ln(2)2^t  +(1/t)  ⇒f^′ (t) =−((ln(2)2^t  +(1/t))/((2^t  +lnt)^2 ))<0 ⇒f decrease on [1,+∞[  so S_n is  a alternate serie ⇒ S_n  converges.

$${we}\:{have}\:{S}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {u}_{{n}} \:\:{with}\:{u}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} \:+{ln}\left({n}\right)} \\ $$$${we}\:{have}\:{u}_{{n}} >\mathrm{0}\:\:\:{and}\:{u}_{{n}} \rightarrow\mathrm{0}\:\:{let}\:{prove}\:{that}\:\left({u}_{{n}} \right){is}\:{decreasing} \\ $$$${let}\:{f}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{t}} \:+{ln}\left({t}\right)}\:\:{with}\:{t}\geqslant\mathrm{1}\:\:{we}\:{have}\: \\ $$$${f}^{'} \left({t}\right)\:=−\frac{\left(\mathrm{2}^{{t}} \:+{ln}\left({t}\right)\right)^{'} }{\left(\mathrm{2}^{{t}} \:+{lnt}\right)^{\mathrm{2}} }\:\:\:\:{and}\:\:\left(\mathrm{2}^{{t}} \:+{lnt}\right)^{'} \:=\left({e}^{{tln}\mathrm{2}} +{lnt}\right)^{'} ={ln}\left(\mathrm{2}\right)\mathrm{2}^{{t}} \:+\frac{\mathrm{1}}{{t}} \\ $$$$\Rightarrow{f}^{'} \left({t}\right)\:=−\frac{{ln}\left(\mathrm{2}\right)\mathrm{2}^{{t}} \:+\frac{\mathrm{1}}{{t}}}{\left(\mathrm{2}^{{t}} \:+{lnt}\right)^{\mathrm{2}} }<\mathrm{0}\:\Rightarrow{f}\:{decrease}\:{on}\:\left[\mathrm{1},+\infty\left[\:\:{so}\:{S}_{{n}} {is}\right.\right. \\ $$$${a}\:{alternate}\:{serie}\:\Rightarrow\:{S}_{{n}} \:{converges}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com