Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 150059 by mathdanisur last updated on 09/Aug/21

Σ_(n=1) ^∞  (1/(n∙(2n + 1))) = ?

$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}\centerdot\left(\mathrm{2n}\:+\:\mathrm{1}\right)}\:=\:? \\ $$

Answered by Kamel last updated on 09/Aug/21

S=2Σ_(n=1) ^(+∞) ((1/(2n))−(1/(2n+1)))=2Σ_(n=1) ^(+∞) (((−1)^n )/n)+2=2(1−Ln(2))

$${S}=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}+\mathrm{2}=\mathrm{2}\left(\mathrm{1}−{Ln}\left(\mathrm{2}\right)\right) \\ $$

Commented by mathdanisur last updated on 09/Aug/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

Answered by Ar Brandon last updated on 09/Aug/21

S=Σ_(n=1) ^∞ (1/(n(2n+1)))=Σ_(n=1) ^∞ ((1/n)−(2/(2n+1)))     =Σ_(n=1) ^∞ ((1/n)−(1/(n+(1/2))))=ψ((3/2))−ψ(1)     =2+ψ((1/2))−ψ(1)=2−γ−2ln2+γ=2−ln4

$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{2}}}\right)=\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right) \\ $$$$\:\:\:=\mathrm{2}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)=\mathrm{2}−\gamma−\mathrm{2ln2}+\gamma=\mathrm{2}−\mathrm{ln4} \\ $$

Commented by Ar Brandon last updated on 09/Aug/21

ψ(x)=−γ+Σ_(n=0) ^∞ ((1/(n+1))−(1/(n+x))) = ((Γ′(x))/(Γ(x)))  ψ(x+1)=(1/x)+ψ(x) , ψ((1/2))=−γ−2ln2  ψ((r/m))=−γ−ln(2m)−(π/2)cot(((rπ)/m))+2Σ_(n=1) ^(∣((m−1)/2)∣) cos(((2πnr)/m))ln sin(((πn)/m))  ψ(z+1)=−γ+Σ_(n=2) ^∞ ζ(n)z^(n−1)   ψ(x)=−γ+∫_0 ^1 ((1−t^(x−1) )/(1−t))dt  ψ′(x)=−∫_0 ^1 ((t^(x−1) ln(t))/(1−t))dt  ψ(x)−ψ(1−x)=−πcot(πx)  ψ^((1)) (x)+ψ^((1)) (1−x)=π^2 (1+cot^2 (πx))  ψ^((2)) (x)−ψ^((2)) (1−x)=−2π^3 (1+cot^2 (πx))cot(πx)

$$\psi\left(\mathrm{x}\right)=−\gamma+\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{x}}\right)\:=\:\frac{\Gamma'\left(\mathrm{x}\right)}{\Gamma\left(\mathrm{x}\right)} \\ $$$$\psi\left(\mathrm{x}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{x}}+\psi\left(\mathrm{x}\right)\:,\:\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\gamma−\mathrm{2ln2} \\ $$$$\psi\left(\frac{\mathrm{r}}{\mathrm{m}}\right)=−\gamma−\mathrm{ln}\left(\mathrm{2m}\right)−\frac{\pi}{\mathrm{2}}\mathrm{cot}\left(\frac{\mathrm{r}\pi}{\mathrm{m}}\right)+\mathrm{2}\underset{\mathrm{n}=\mathrm{1}} {\overset{\mid\frac{\mathrm{m}−\mathrm{1}}{\mathrm{2}}\mid} {\sum}}\mathrm{cos}\left(\frac{\mathrm{2}\pi\mathrm{nr}}{\mathrm{m}}\right)\mathrm{ln}\:\mathrm{sin}\left(\frac{\pi\mathrm{n}}{\mathrm{m}}\right) \\ $$$$\psi\left({z}+\mathrm{1}\right)=−\gamma+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\zeta\left({n}\right){z}^{{n}−\mathrm{1}} \\ $$$$\psi\left(\mathrm{x}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\mathrm{t}^{\mathrm{x}−\mathrm{1}} }{\mathrm{1}−\mathrm{t}}\mathrm{dt} \\ $$$$\psi'\left(\mathrm{x}\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{t}^{\mathrm{x}−\mathrm{1}} \mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}−\mathrm{t}}\mathrm{dt} \\ $$$$\psi\left(\mathrm{x}\right)−\psi\left(\mathrm{1}−\mathrm{x}\right)=−\pi\mathrm{cot}\left(\pi\mathrm{x}\right) \\ $$$$\psi^{\left(\mathrm{1}\right)} \left(\mathrm{x}\right)+\psi^{\left(\mathrm{1}\right)} \left(\mathrm{1}−\mathrm{x}\right)=\pi^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cot}^{\mathrm{2}} \left(\pi\mathrm{x}\right)\right) \\ $$$$\psi^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)−\psi^{\left(\mathrm{2}\right)} \left(\mathrm{1}−\mathrm{x}\right)=−\mathrm{2}\pi^{\mathrm{3}} \left(\mathrm{1}+\mathrm{cot}^{\mathrm{2}} \left(\pi\mathrm{x}\right)\right)\mathrm{cot}\left(\pi\mathrm{x}\right) \\ $$

Commented by mathdanisur last updated on 09/Aug/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

Commented by Tawa11 last updated on 09/Aug/21

weldone sir

$$\mathrm{weldone}\:\mathrm{sir} \\ $$

Commented by Tawa11 last updated on 09/Aug/21

sir Ar brandon, please give me the definition of   ψ  that you used.  or books sir. please ...

$$\mathrm{sir}\:\mathrm{Ar}\:\mathrm{brandon},\:\mathrm{please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of}\:\:\:\psi\:\:\mathrm{that}\:\mathrm{you}\:\mathrm{used}. \\ $$$$\mathrm{or}\:\mathrm{books}\:\mathrm{sir}.\:\mathrm{please}\:... \\ $$

Commented by mnjuly1970 last updated on 09/Aug/21

 excellent  Sir Brandon...

$$\:{excellent}\:\:\mathrm{S}{ir}\:{Brandon}... \\ $$

Commented by Ar Brandon last updated on 09/Aug/21

Thank you Sir !

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}\:! \\ $$

Commented by Ar Brandon last updated on 09/Aug/21

They are part of your teachings  😃

$$\mathrm{They}\:\mathrm{are}\:\mathrm{part}\:\mathrm{of}\:\mathrm{your}\:\mathrm{teachings} \\ $$😃

Commented by Tawa11 last updated on 09/Aug/21

Wow, thanks sir. God bless you. I really appreciate.

$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by Ar Brandon last updated on 09/Aug/21

My pleasure

Terms of Service

Privacy Policy

Contact: info@tinkutara.com