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Question Number 63101 by mathmax by abdo last updated on 29/Jun/19

calculate S =(1/(1×2)) +(1/(3×4)) +(1/(5×6)) +.....

$${calculate}\:\:{S}\:=\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{6}}\:+..... \\ $$

Commented by mathmax by abdo last updated on 29/Jun/19

let try another way we have (d/dx)ln(1+x) =(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n with ∣x∣<1 ⇒ ln(1+x) =Σ_(n=0) ^∞ (((−1)^n x^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n) ⇒ lim_(x→1) ln(1+x)=ln(2) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) =1−(1/2) +(1/3) −(1/4) +(1/5) −(1/6) +...= =(1−(1/2))+((1/3)−(1/4))+((1/5) −(1/6))+...=(1/(1×2)) +(1/(3×4)) +(1/(5×6)) +....=S ⇒ S =ln(2).

$${let}\:{try}\:{another}\:{way}\:{we}\:{have}\:\frac{{d}}{{dx}}{ln}\left(\mathrm{1}+{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{{n}} \:{with}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {x}^{{n}} }{{n}}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} {ln}\left(\mathrm{1}+{x}\right)={ln}\left(\mathrm{2}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:−\frac{\mathrm{1}}{\mathrm{6}}\:+...= \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\left(\frac{\mathrm{1}}{\mathrm{5}}\:−\frac{\mathrm{1}}{\mathrm{6}}\right)+...=\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{6}}\:+....={S}\:\Rightarrow \\ $$$${S}\:={ln}\left(\mathrm{2}\right). \\ $$

Answered by mr W last updated on 29/Jun/19

S =(1/(1×2)) +(1/(3×4)) +(1/(5×6)) +..... >0 S=((1/1)−(1/2))+((1/3)−(1/4))+((1/5)−(1/6))+ S=((1/1)+(1/3)+(1/5)+...)−((1/2)+(1/4)+(1/6)+...) S=((1/1)+(1/2)+(1/3)+(1/4)+(1/5)+(1/6)+...)−2((1/2)+(1/4)+(1/6)+...) S=((1/1)+(1/2)+(1/3)+(1/4)+(1/5)+(1/6)+...)−((1/1)+(1/2)+(1/3)+...) S=0 ?

$${S}\:=\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{6}}\:+.....\:>\mathrm{0} \\ $$$${S}=\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{6}}\right)+ \\ $$$${S}=\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+...\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+...\right) \\ $$$${S}=\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+...\right)−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+...\right) \\ $$$${S}=\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+...\right)−\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...\right) \\ $$$${S}=\mathrm{0}\:? \\ $$

Commented by MJS last updated on 29/Jun/19

I seem to remember that it′s not allowed to rearrange in this way S=(1/2)+s ⇒ S>(1/2) S=(7/(12))+s ⇒ S>(7/(12)) ...

$$\mathrm{I}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{remember}\:\mathrm{that}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{allowed} \\ $$$$\mathrm{to}\:\mathrm{rearrange}\:\mathrm{in}\:\mathrm{this}\:\mathrm{way} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}+{s}\:\Rightarrow\:{S}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${S}=\frac{\mathrm{7}}{\mathrm{12}}+{s}\:\Rightarrow\:{S}>\frac{\mathrm{7}}{\mathrm{12}} \\ $$$$... \\ $$

Commented by turbo msup by abdo last updated on 29/Jun/19

we hsve S=Σa_i and a_i >0 how can S be0....

$${we}\:{hsve}\:{S}=\Sigma{a}_{{i}} \:\:\:{and}\:{a}_{{i}} \:>\mathrm{0}\:{how} \\ $$$${can}\:{S}\:{be}\mathrm{0}.... \\ $$

Commented by turbo msup by abdo last updated on 29/Jun/19

we hsve S=Σa_i and a_i >0 how can S be0....

$${we}\:{hsve}\:{S}=\Sigma{a}_{{i}} \:\:\:{and}\:{a}_{{i}} \:>\mathrm{0}\:{how} \\ $$$${can}\:{S}\:{be}\mathrm{0}.... \\ $$

Commented by mathmax by abdo last updated on 29/Jun/19

sir mrw your begining is correct but the answer is not 0 let complete the work we have S =lim_(n→+∞) ( Σ_(k=0) ^n (1/(2k+1)) −(1/2)Σ_(k=1) ^n (1/k)) we have Σ_(k=1) ^n (1/k) =H_n Σ_(k=0) ^n (1/(2k+1)) =1+(1/3) +(1/5) +....+(1/(2n+1)) =1+(1/2) +(1/3) +(1/4) +(1/5) +....+(1/(2n)) +(1/(2n+1)) −(1/2) −(1/4) −....−(1/(2n)) =H_(2n+1) −(1/2) H_n ⇒ Σ_(k=0) ^n (1/(2k+1)) −(1/2)Σ_(k=1) ^n (1/k) =H_(2n+1) −H_n =ln(2n+1)+γ +o((1/n))−ln(n)−γ +o((1/n)) =ln(((2n+1)/n))+o((1/n))→ln(2) (n→+∞) so (1/(1×2)) +(1/(3×4)) +(1/(5×6)) +.....=ln(2).(γ is the constant of euler)

$${sir}\:{mrw}\:{your}\:{begining}\:{is}\:{correct}\:\:{but}\:{the}\:{answer}\:{is}\:{not}\:\mathrm{0}\:{let}\:{complete}\:{the} \\ $$$${work}\:{we}\:{have}\:{S}\:={lim}_{{n}\rightarrow+\infty} \left(\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\right) \\ $$$${we}\:{have}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+....+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+....+\frac{\mathrm{1}}{\mathrm{2}{n}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:−....−\frac{\mathrm{1}}{\mathrm{2}{n}}\:={H}_{\mathrm{2}{n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{\mathrm{2}{n}+\mathrm{1}} −{H}_{{n}} ={ln}\left(\mathrm{2}{n}+\mathrm{1}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)−{ln}\left({n}\right)−\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$={ln}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{{n}}\right)+{o}\left(\frac{\mathrm{1}}{{n}}\right)\rightarrow{ln}\left(\mathrm{2}\right)\:\left({n}\rightarrow+\infty\right)\:{so} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{6}}\:+.....={ln}\left(\mathrm{2}\right).\left(\gamma\:{is}\:{the}\:{constant}\:{of}\:{euler}\right) \\ $$

Commented by mr W last updated on 29/Jun/19

thanks sir! i learnt something new.

$${thanks}\:{sir}!\:{i}\:{learnt}\:{something}\:{new}. \\ $$

Commented by mathmax by abdo last updated on 29/Jun/19

you are always welcome.

$${you}\:{are}\:{always}\:{welcome}. \\ $$

Answered by MJS last updated on 29/Jun/19

it seems to be S=ln 2 but I cannot properly show it

$$\mathrm{it}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be} \\ $$$${S}=\mathrm{ln}\:\mathrm{2} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{properly}\:\mathrm{show}\:\mathrm{it} \\ $$

Answered by naka3546 last updated on 29/Jun/19

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