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Question Number 63574 by Tawa1 last updated on 05/Jul/19

prove that Σ_(k = 1) ^∞ (1/(k(2k + 1))) = 2 − 2ln(2)

$$\mathrm{prove}\:\mathrm{that}\:\:\:\underset{\mathrm{k}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{2k}\:+\:\mathrm{1}\right)}\:\:=\:\:\mathrm{2}\:−\:\mathrm{2ln}\left(\mathrm{2}\right) \\ $$

Commented by mathmax by abdo last updated on 05/Jul/19

the H_n method let S_n =Σ_(k=1) ^n (1/(k(2k+1))) ⇒(1/2) S_n =Σ_(k=1) ^n (1/(2k(2k+1))) =Σ_(k=1) ^n {(1/(2k)) −(1/(2k+1))} =(1/2) Σ_(k=1) ^n (1/k) −Σ_(k=1) ^n (1/(2k+1)) we have Σ_(k=1) ^n (1/k) =H_n Σ_(k=1) ^n (1/(2k+1)) =(1/3) +(1/5) +......+(1/(2n+1)) =1+(1/2) +(1/3) +(1/4) +....+(1/(2n)) +(1/(2n+1)) −1−(1/2)−(1/4)−....−(1/(2n)) =H_(2n+1) −1−(1/2) H_n ⇒ (1/2) S_n =(1/2) H_n −H_(2n+1) +1+(1/2)H_n =H_n −H_(2n+1) +1 =ln(n)+γ +o((1/n))−ln(2n+1)−γ +o((1/n)) +1 =ln((n/(2n+1))) +o((1/n))+1 ⇒ S_n =2ln((n/(2n+1))) +2 ⇒ lim_(n→+∞) S_n =2ln((1/2))+2 =−2ln(2) +2 .the result is proved.

$${the}\:{H}_{{n}} \:{method}\:\:{let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}\left(\mathrm{2}{k}+\mathrm{1}\right)}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \left\{\frac{\mathrm{1}}{\mathrm{2}{k}}\:−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:\:{we}\:{have} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+......+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+....+\frac{\mathrm{1}}{\mathrm{2}{n}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−....−\frac{\mathrm{1}}{\mathrm{2}{n}}\:={H}_{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} −{H}_{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} ={H}_{{n}} −{H}_{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1} \\ $$$$={ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)−{ln}\left(\mathrm{2}{n}+\mathrm{1}\right)−\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:+\mathrm{1} \\ $$$$={ln}\left(\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)+\mathrm{1}\:\Rightarrow\:{S}_{{n}} =\mathrm{2}{ln}\left(\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}\right)\:+\mathrm{2}\:\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\mathrm{2}{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2}\:=−\mathrm{2}{ln}\left(\mathrm{2}\right)\:+\mathrm{2}\:.{the}\:{result}\:{is}\:{proved}. \\ $$

Commented by Prithwish sen last updated on 05/Jul/19

=Σ [(1/k) −(2/(2k+1))] =(1+(1/2) +(1/3)+ ..........∞) −2((1/3) + (1/5) + ......∞) = (1+(1/2) − (1/3) + (1/4) − (1/5) + (1/6) − .......∞) = 2−(1−(1/2) + (1/3) − (1/4) + .......∞) =2 − ln(2) I. can′t find out my mistake. Please help anyone

$$=\Sigma\:\left[\frac{\mathrm{1}}{\mathrm{k}}\:−\frac{\mathrm{2}}{\mathrm{2k}+\mathrm{1}}\right] \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}+\:..........\infty\right)\:−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{5}}\:+\:......\infty\right) \\ $$$$=\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{6}}\:−\:.......\infty\right) \\ $$$$=\:\mathrm{2}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:.......\infty\right) \\ $$$$=\mathrm{2}\:−\:\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{I}.\:\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\mathrm{out}\:\mathrm{my}\:\mathrm{mistake}.\:\mathrm{Please}\:\mathrm{help}\:\mathrm{anyone} \\ $$

Commented by Tawa1 last updated on 05/Jul/19

God bless you sirs

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sirs} \\ $$

Commented by mathmax by abdo last updated on 05/Jul/19

method of series let S = Σ_(n=1) ^∞ (1/(n(2n+1))) and S(x)=Σ_(n=1) ^∞ (x^(2n+1) /(n(2n+1))) we have S=S(1) we have (dS/dx)(x)=Σ_(n=1) ^∞ (x^(2n) /n) =w(x^2 ) with w(t)=Σ_(n=1) ^∞ (t^n /n) we have w^′ (t) =Σ_(n=1) ^∞ t^(n−1) =Σ_(n=0) ^∞ t^n =(1/(1−t)) ⇒w(t)=−ln∣1−t∣ +c c=w(0)=0 ⇒(dS/dx)(x)=−ln(1−x^2 ) (we take −1<x<1) ⇒ S(x)=−∫_0 ^x ln(1−t^2 )dt +λ (λ=S(0)=0) ⇒S(x)=−∫_0 ^x ln(1−t^2 )dt ⇒ S =−∫_0 ^1 ln(1−t^2 )dt =− ∫_0 ^1 ln(1−t) −∫_0 ^1 ln(1+t)dt ∫_0 ^1 ln(1−t)dt =_(1−t =u) ∫_0 ^1 ln(u)du =[uln(u)−u]_0 ^1 =−1 ∫_0 ^1 ln(1+t)dt =_(1+t =u) ∫_1 ^2 ln(u)du =[uln(u)−u]_1 ^2 =2ln(2)−2+1 ⇒ S =1−2ln(2)+2−1 =2−2ln(2).

$${method}\:{of}\:{series}\:{let}\:{S}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:\:\:{and}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\:{we}\:{have}\:{S}={S}\left(\mathrm{1}\right)\:\:{we}\:{have}\:\frac{{dS}}{{dx}}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{{n}}\:={w}\left({x}^{\mathrm{2}} \right)\:{with}\:{w}\left({t}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}} }{{n}} \\ $$$${we}\:{have}\:{w}^{'} \left({t}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{t}^{{n}−\mathrm{1}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{t}}\:\Rightarrow{w}\left({t}\right)=−{ln}\mid\mathrm{1}−{t}\mid\:+{c} \\ $$$${c}={w}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\frac{{dS}}{{dx}}\left({x}\right)=−{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:\:\:\:\:\left({we}\:{take}\:\:−\mathrm{1}<{x}<\mathrm{1}\right)\:\Rightarrow \\ $$$${S}\left({x}\right)=−\int_{\mathrm{0}} ^{{x}} {ln}\left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}\:+\lambda\:\:\:\:\:\left(\lambda={S}\left(\mathrm{0}\right)=\mathrm{0}\right)\:\Rightarrow{S}\left({x}\right)=−\int_{\mathrm{0}} ^{{x}} {ln}\left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}\:\Rightarrow \\ $$$${S}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}\:=−\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{t}\right)\:−\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{t}\right){dt}\:=_{\mathrm{1}−{t}\:={u}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({u}\right){du}\:=\left[{uln}\left({u}\right)−{u}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=−\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}\right){dt}\:=_{\mathrm{1}+{t}\:={u}} \:\:\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:{ln}\left({u}\right){du}\:=\left[{uln}\left({u}\right)−{u}\right]_{\mathrm{1}} ^{\mathrm{2}} \:=\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{2}+\mathrm{1}\:\Rightarrow \\ $$$${S}\:=\mathrm{1}−\mathrm{2}{ln}\left(\mathrm{2}\right)+\mathrm{2}−\mathrm{1}\:=\mathrm{2}−\mathrm{2}{ln}\left(\mathrm{2}\right). \\ $$$$ \\ $$

Commented by Tawa1 last updated on 05/Jul/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 05/Jul/19

Commented by Tawa1 last updated on 05/Jul/19

Please help sir

$$\mathrm{Please}\:\mathrm{help}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 05/Jul/19

I appreciate your time sir

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir} \\ $$

Commented by Prithwish sen last updated on 06/Jul/19

What is the answer ?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:? \\ $$

Commented by Tawa1 last updated on 06/Jul/19

I don′t know the answer sir. But you can help sir.

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{sir}.\:\mathrm{But}\:\mathrm{you}\:\mathrm{can}\:\mathrm{help}\:\mathrm{sir}. \\ $$

Commented by Prithwish sen last updated on 06/Jul/19

a_(n+1) =(√(1+n(√(1+(n−1)(√(1+(n−2)(√(1+....)))))))) Now, n−1 = (√((n−1)^2 )) = (√(1 + n(n−2))) = (√(1+n(√((n−2)^2 )))) = (√(1+n(√(1+ (n−1)(n−2))))) = (√(1+n(√(1+(n−1)(√(1+(n−2)(√(1+..)))))))) as n→∞ a_(n+1) → (n−1)⇒a_n → (n−2) ⇒ (a_n /n) → (1−(2/n))→ 1 Ω = lim_(n→∞) [(1/n){ (1/((n + (a_n /n) )))+ (2/((n + 2.(a_n /n) ))) + .....+ (n/((n + n.(a_n /n) )))}]^(n^(1/n) −1) = lim_(n→∞ ) [(1/n){(l/(n+1)) +(2/(n+2)) + .......... + (n/(n+n)) }]^n^((1/n) −1) = lim_(n→∞) [(1/n){((1/n)/(1+(1/n))) + ((2/n)/(1+ (2/n))) + .....+ ((n/n)/(1 + (n/n))) }]^n^((1/n) −1) taking log ln(Ω) = lim_(n→∞) (n^(1/n) −1) ln [ lim_(n→∞) (1/n) Σ_(r=1) ^n ((r/n)/(1+(r/n))) ] = lim_(n→∞) (n^(1/n) −1) ln[∫_0 ^1 ((xdx)/(1+x))] = lim_(n→∞ ) (n^(1/n) −1)ln(1−ln2) Now as n→∞ lnΩ → 0 ⇒ Ω→1 I don′t know whether it is corrector not.Please give your valuable suggestion.

$$\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\sqrt{\mathrm{1}+\mathrm{n}\sqrt{\mathrm{1}+\left(\mathrm{n}−\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{n}−\mathrm{2}\right)\sqrt{\mathrm{1}+....}}}} \\ $$$$\mathrm{Now},\:\mathrm{n}−\mathrm{1}\:=\:\sqrt{\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \:}\:=\:\sqrt{\mathrm{1}\:+\:\mathrm{n}\left(\mathrm{n}−\mathrm{2}\right)} \\ $$$$\:\:=\:\sqrt{\mathrm{1}+\mathrm{n}\sqrt{\left(\mathrm{n}−\mathrm{2}\right)^{\mathrm{2}} }}\:=\:\sqrt{\mathrm{1}+\mathrm{n}\sqrt{\mathrm{1}+\:\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)}} \\ $$$$\:\:\:=\:\sqrt{\mathrm{1}+\mathrm{n}\sqrt{\mathrm{1}+\left(\mathrm{n}−\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{n}−\mathrm{2}\right)\sqrt{\mathrm{1}+..}}}} \\ $$$$\mathrm{as}\:\mathrm{n}\rightarrow\infty\:\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \rightarrow\:\left(\mathrm{n}−\mathrm{1}\right)\Rightarrow\mathrm{a}_{\mathrm{n}} \rightarrow\:\left(\mathrm{n}−\mathrm{2}\right)\:\Rightarrow\:\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{n}}\:\rightarrow\:\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{n}}\right)\rightarrow\:\mathrm{1} \\ $$$$\Omega\:=\:\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\left[\frac{\mathrm{1}}{\mathrm{n}}\left\{\:\frac{\mathrm{1}}{\left(\mathrm{n}\:+\:\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{n}}\:\right)}+\:\frac{\mathrm{2}}{\left(\mathrm{n}\:+\:\mathrm{2}.\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{n}}\:\right)}\:\:+\:\:.....+\:\frac{\mathrm{n}}{\left(\mathrm{n}\:+\:\mathrm{n}.\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{n}}\:\right)}\right\}\right]^{\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{n}}} −\mathrm{1}} \\ $$$$=\:\mathrm{li}\underset{\mathrm{n}\rightarrow\infty\:\:} {\mathrm{m}}\:\left[\frac{\mathrm{1}}{\mathrm{n}}\left\{\frac{\mathrm{l}}{\mathrm{n}+\mathrm{1}}\:+\frac{\mathrm{2}}{\mathrm{n}+\mathrm{2}}\:+\:..........\:+\:\frac{\mathrm{n}}{\mathrm{n}+\mathrm{n}}\:\right\}\right]^{\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{n}}\:−\mathrm{1}} } \\ $$$$=\:\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\left[\frac{\mathrm{1}}{\mathrm{n}}\left\{\frac{\frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}\:+\:\frac{\frac{\mathrm{2}}{\mathrm{n}}}{\mathrm{1}+\:\frac{\mathrm{2}}{\mathrm{n}}}\:+\:.....+\:\frac{\frac{\mathrm{n}}{\mathrm{n}}}{\mathrm{1}\:+\:\frac{\mathrm{n}}{\mathrm{n}}}\:\right\}\right]^{\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{n}}\:−\mathrm{1}} } \\ $$$$\mathrm{taking}\:\mathrm{log} \\ $$$$\mathrm{ln}\left(\Omega\right)\:=\:\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\left(\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{n}}} −\mathrm{1}\right)\:\mathrm{ln}\:\left[\:\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{n}}\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\frac{\mathrm{r}}{\mathrm{n}}}{\mathrm{1}+\frac{\mathrm{r}}{\mathrm{n}}}\:\right] \\ $$$$\:=\:\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\left(\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{n}}} −\mathrm{1}\right)\:\mathrm{ln}\left[\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{xdx}}{\mathrm{1}+\mathrm{x}}\right] \\ $$$$=\:\mathrm{lim}_{\mathrm{n}\rightarrow\infty\:} \left(\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{n}}} −\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}−\mathrm{ln2}\right) \\ $$$$\mathrm{Now}\:\mathrm{as}\:\mathrm{n}\rightarrow\infty\:\mathrm{ln}\Omega\:\rightarrow\:\mathrm{0}\:\Rightarrow\:\Omega\rightarrow\mathrm{1} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{whether}\:\mathrm{it}\:\mathrm{is}\:\mathrm{corrector}\:\mathrm{not}.\mathrm{Please} \\ $$$$\mathrm{give}\:\mathrm{your}\:\mathrm{valuable}\:\mathrm{suggestion}. \\ $$

Commented by Prithwish sen last updated on 06/Jul/19

thanks abdo sir

$$\mathrm{thanks}\:\mathrm{abdo}\:\mathrm{sir} \\ $$

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