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Question Number 132715 by Dwaipayan Shikari last updated on 16/Feb/21

(1/1^3 )−(1/2^3 )+(1/4^3 )−(1/5^3 )+(1/7^3 )−(1/8^3 )+...

$$\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{8}^{\mathrm{3}} }+... \\ $$

Answered by Olaf last updated on 16/Feb/21

Let (a_n )_(n∈N) = {1, 2, 4, 5, 7, 8...} a_n = (1/4)[6n+3+(−1)^n ] S = Σ_(n=0) ^∞ (((−1)^n )/a_n ^3 ) S = Σ_(n=0) ^∞ (((−1)^n 4^3 )/([6n+3+(−1)^n ]^3 )) S = Σ_(p=0) ^∞ (((−1)^(2p) 4^3 )/([6(2p)+3+(−1)^(2p) ]^3 )) + Σ_(p=0) ^∞ (((−1)^(2p+1) 4^3 )/([6(2p+1)+3+(−1)^(2p+1) ]^3 )) S = Σ_(p=0) ^∞ (4^3 /((12p+4)^3 ))−Σ_(p=0) ^∞ (4^3 /((12p+8)^3 )) S = Σ_(p=0) ^∞ (1/((3p+1)^3 ))−Σ_(p=0) ^∞ (1/((3p+2)^3 )) S = −((Ψ(2,(1/3)))/(54))+((Ψ(2,(2/3)))/(54)) S = ((4π^3 (√3))/(243)) S ≈ 0,8840238122

$$\mathrm{Let}\:\left({a}_{{n}} \right)_{{n}\in\mathbb{N}} \:=\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{4},\:\mathrm{5},\:\mathrm{7},\:\mathrm{8}...\right\} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{6}{n}+\mathrm{3}+\left(−\mathrm{1}\right)^{{n}} \right] \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{a}_{{n}} ^{\mathrm{3}} } \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{4}^{\mathrm{3}} }{\left[\mathrm{6}{n}+\mathrm{3}+\left(−\mathrm{1}\right)^{{n}} \right]^{\mathrm{3}} } \\ $$$$ \\ $$$$\mathrm{S}\:=\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{p}} \mathrm{4}^{\mathrm{3}} }{\left[\mathrm{6}\left(\mathrm{2}{p}\right)+\mathrm{3}+\left(−\mathrm{1}\right)^{\mathrm{2}{p}} \right]^{\mathrm{3}} } \\ $$$$+\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{p}+\mathrm{1}} \mathrm{4}^{\mathrm{3}} }{\left[\mathrm{6}\left(\mathrm{2}{p}+\mathrm{1}\right)+\mathrm{3}+\left(−\mathrm{1}\right)^{\mathrm{2}{p}+\mathrm{1}} \right]^{\mathrm{3}} } \\ $$$$ \\ $$$$\mathrm{S}\:=\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{\mathrm{3}} }{\left(\mathrm{12}{p}+\mathrm{4}\right)^{\mathrm{3}} }−\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{\mathrm{3}} }{\left(\mathrm{12}{p}+\mathrm{8}\right)^{\mathrm{3}} } \\ $$$$\mathrm{S}\:=\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{p}+\mathrm{1}\right)^{\mathrm{3}} }−\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{p}+\mathrm{2}\right)^{\mathrm{3}} } \\ $$$$\mathrm{S}\:=\:−\frac{\Psi\left(\mathrm{2},\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{54}}+\frac{\Psi\left(\mathrm{2},\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{54}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{4}\pi^{\mathrm{3}} \sqrt{\mathrm{3}}}{\mathrm{243}} \\ $$$$\mathrm{S}\:\approx\:\mathrm{0},\mathrm{8840238122} \\ $$

Commented by Dwaipayan Shikari last updated on 16/Feb/21

Great sir!

$${Great}\:{sir}! \\ $$

Commented by Dwaipayan Shikari last updated on 16/Feb/21

(π/2)tan((π/2)x)=(1/(1−x))−(1/(1+x))+(1/(3−x))−(1/(3+x))+... (π^2 /4)sec^2 ((π/2)x)=(1/((1−x)^2 ))+(1/((1+x)^2 ))+(1/((3−x)^2 ))+(1/((3+x)^2 ))+... (π^3 /8)sec^2 ((π/2)x)tan((π/2)x)=(1/((1−x)^3 ))−(1/((1+x)^3 ))+... x=(1/3) (π^3 /8).(4/3).(1/( (√3)))=(3^3 /2^3 )−(3^3 /4^3 )+(3^3 /8^3 )−(3^3 /(10^3 ))+... =(π^3 /(27.6(√3)))=(1/2^3 )(1−(1/2^3 )+(1/4^3 )−(1/5^3 )+...) ⇒((4π^3 )/(81(√3)))=1−(1/2^3 )+(1/4^3 )−(1/5^3 )+...

$$\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}−{x}}−\frac{\mathrm{1}}{\mathrm{3}+{x}}+... \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{3}+{x}\right)^{\mathrm{2}} }+... \\ $$$$\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{x}\right){tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }+... \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\pi^{\mathrm{3}} }{\mathrm{8}}.\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} }−\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{8}^{\mathrm{3}} }−\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{10}^{\mathrm{3}} }+... \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{27}.\mathrm{6}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+...\right) \\ $$$$\Rightarrow\frac{\mathrm{4}\pi^{\mathrm{3}} }{\mathrm{81}\sqrt{\mathrm{3}}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+... \\ $$

Commented by mnjuly1970 last updated on 16/Feb/21

please prove this nice identity mr dwaipayan.. (π/2)tan((π/2)(x))=(1/(1−x))−(1/(1+x)) +(1/(3+x))−(1/(3−x))... of course i think it can be deduced by weiresteress gamma formula.. Γ(x)=(e^(−γx) /x)Π_(k=1) ^∞ (e^(x/k) /(1+(x/k))) and its corallary ((sin(πx))/(πx))=Π_(k=1) ^∞ (1−(x^2 /k^2 )) ..

$${please}\:{prove}\:{this}\:{nice}\:{identity} \\ $$$${mr}\:{dwaipayan}.. \\ $$$$\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi}{\mathrm{2}}\left({x}\right)\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}\:+\frac{\mathrm{1}}{\mathrm{3}+{x}}−\frac{\mathrm{1}}{\mathrm{3}−{x}}... \\ $$$${of}\:{course}\:{i}\:{think} \\ $$$$\:{it}\:{can}\:{be}\:{deduced}\:{by} \\ $$$$\:{weiresteress}\:{gamma}\:{formula}.. \\ $$$$\:\Gamma\left({x}\right)=\frac{{e}^{−\gamma{x}} }{{x}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{{e}^{\frac{{x}}{{k}}} }{\mathrm{1}+\frac{{x}}{{k}}} \\ $$$${and}\:{its}\:{corallary} \\ $$$$\:\frac{{sin}\left(\pi{x}\right)}{\pi{x}}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} }\right)\:.. \\ $$

Commented by Dwaipayan Shikari last updated on 16/Feb/21

cosx=(1−((2x)/π))(1+((2x)/π))(1−((2x)/(3π)))(1+((2x)/(3π)))(1−((2x)/(5π)))(1+((2x)/(5π)))... log(cosx)=log(1−((2x)/π))+log(1+((2x)/π))+... Differentiating both sides respect to x −tanx=(((−2)/π)/(1−((2x)/π)))+((2/π)/(1+((2x)/π)))−((2/(3π))/(1−((2x)/(3π))))+.. tanx=(1/((π/2)−x))−(1/((π/2)+x))+(1/(((3π)/2)−x))−(1/(((3π)/2)+x))+... (π/2)tan((π/2)x)=(1/(1−x))−(1/(1+x))+(1/(3−x))−(1/(3+x))+(1/(5−x))−(1/(5+x))+...

$${cosx}=\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\pi}\right)\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{5}\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{5}\pi}\right)... \\ $$$${log}\left({cosx}\right)={log}\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\pi}\right)+{log}\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\pi}\right)+... \\ $$$${Differentiating}\:{both}\:{sides}\:{respect}\:{to}\:{x} \\ $$$$−{tanx}=\frac{\frac{−\mathrm{2}}{\pi}}{\mathrm{1}−\frac{\mathrm{2}{x}}{\pi}}+\frac{\frac{\mathrm{2}}{\pi}}{\mathrm{1}+\frac{\mathrm{2}{x}}{\pi}}−\frac{\frac{\mathrm{2}}{\mathrm{3}\pi}}{\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}\pi}}+.. \\ $$$${tanx}=\frac{\mathrm{1}}{\frac{\pi}{\mathrm{2}}−{x}}−\frac{\mathrm{1}}{\frac{\pi}{\mathrm{2}}+{x}}+\frac{\mathrm{1}}{\frac{\mathrm{3}\pi}{\mathrm{2}}−{x}}−\frac{\mathrm{1}}{\frac{\mathrm{3}\pi}{\mathrm{2}}+{x}}+... \\ $$$$\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}−{x}}−\frac{\mathrm{1}}{\mathrm{3}+{x}}+\frac{\mathrm{1}}{\mathrm{5}−{x}}−\frac{\mathrm{1}}{\mathrm{5}+{x}}+... \\ $$

Commented by mnjuly1970 last updated on 16/Feb/21

grateful mr Dwaipayan thanks lot...

$${grateful}\:{mr}\:{Dwaipayan} \\ $$$${thanks}\:{lot}... \\ $$

Answered by mnjuly1970 last updated on 16/Feb/21

solution: S is convergent... rearrangement... S=((1/1^3 )+(1/4^3 )+(1/7^3 )+....)+ ((1/2^3 )+(1/5^3 )+(1/8^3 )+...) =Σ_(n=0) ^∞ (1/((3n+1)^3 )) −Σ_(n=1) ^∞ (1/((3n−1)^3 )) =(1/(27)){Σ_(n=0 ) ^∞ (1/((n+(1/3))^3 ))−Σ_(n=0) ^∞ (1/((n+(2/3) )^3 ))} note :: (1):: 𝛙(z)−𝛙(1−z)=−𝛑cot(πz) (2): : 𝛙′(z)+𝛙′(1−z)=𝛑^2 (1+cot^2 (πz)) (3) :: 𝛙′′(z)−𝛙′′(1−z)=−2𝛑^3 (1+cot^2 (𝛑z))cot(𝛑z) 𝛙′′((1/3))−𝛙′′((2/3))=−2π^3 (1+(1/3))((1/( (√3)))) 𝛙′′((1/3))−𝛙′′((2/3))=((−8π^3 )/(3(√3))) .... S=(1/(27))(((−𝛙′′((1/3)))/2)+((𝛙′′((2/3)))/2)) =(1/(54))(((8π^3 )/(3(√3))))=((4𝛑^3 )/(81(√3))) ....✓✓ m.n..

$${solution}: \\ $$$$\:\:{S}\:{is}\:{convergent}...\: \\ $$$$\:\:\:{rearrangement}... \\ $$$$\:\:\:{S}=\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }+....\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{8}^{\mathrm{3}} }+...\right) \\ $$$$\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{27}}\left\{\underset{{n}=\mathrm{0}\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} }−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\:\right)^{\mathrm{3}} }\right\} \\ $$$${note}\:::\:\left(\mathrm{1}\right)::\:\:\:\boldsymbol{\psi}\left({z}\right)−\boldsymbol{\psi}\left(\mathrm{1}−{z}\right)=−\boldsymbol{\pi}{cot}\left(\pi{z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right):\::\:\:\boldsymbol{\psi}'\left({z}\right)+\boldsymbol{\psi}'\left(\mathrm{1}−{z}\right)=\boldsymbol{\pi}^{\mathrm{2}} \left(\mathrm{1}+{cot}^{\mathrm{2}} \left(\pi{z}\right)\right) \\ $$$$\:\left(\mathrm{3}\right)\:::\:\:\boldsymbol{\psi}''\left({z}\right)−\boldsymbol{\psi}''\left(\mathrm{1}−{z}\right)=−\mathrm{2}\boldsymbol{\pi}^{\mathrm{3}} \left(\mathrm{1}+{cot}^{\mathrm{2}} \left(\boldsymbol{\pi}{z}\right)\right){cot}\left(\boldsymbol{\pi}{z}\right) \\ $$$$\:\:\:\:\boldsymbol{\psi}''\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\boldsymbol{\psi}''\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=−\mathrm{2}\pi^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\:\:\boldsymbol{\psi}''\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\boldsymbol{\psi}''\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{−\mathrm{8}\pi^{\mathrm{3}} }{\mathrm{3}\sqrt{\mathrm{3}}}\:.... \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{27}}\left(\frac{−\boldsymbol{\psi}''\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{2}}+\frac{\boldsymbol{\psi}''\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{54}}\left(\frac{\mathrm{8}\pi^{\mathrm{3}} }{\mathrm{3}\sqrt{\mathrm{3}}}\right)=\frac{\mathrm{4}\boldsymbol{\pi}^{\mathrm{3}} }{\mathrm{81}\sqrt{\mathrm{3}}}\:\:....\checkmark\checkmark\:{m}.{n}.. \\ $$$$\:\: \\ $$

Answered by Ajetunmobi last updated on 16/Feb/21

let the sum =S seperating the positive term and negative term and factor out minus we get S=Σ_(k=0) ^∞ (1/((3k+1)^3 ))−Σ_(k=0) (1/((3k+2)^3 ))=(1/(27)){Σ_(k=0) (1/((k+(1/3))^3 ))−Σ_(k=0) (1/((k+(2/3))^3 ))} S=(1/(27)){𝛙^((1)) ((1/3))−𝛙^((1)) ((2/3))} Answer Ajetunmobi Abdulqoyyum

$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}}\:=\boldsymbol{\mathrm{S}} \\ $$$$\boldsymbol{\mathrm{seperating}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{negative}}\:\boldsymbol{\mathrm{term}} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{factor}}\:\boldsymbol{\mathrm{out}}\:\boldsymbol{\mathrm{minus}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}} \\ $$$$\boldsymbol{\mathrm{S}}=\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\mathrm{3}} }−\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{3}\boldsymbol{\mathrm{k}}+\mathrm{2}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{27}}\left\{\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{k}}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} }−\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{k}}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }\right\} \\ $$$$\boldsymbol{\mathrm{S}}=\frac{\mathrm{1}}{\mathrm{27}}\left\{\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right\}\:\boldsymbol{\mathrm{Answer}} \\ $$$$\boldsymbol{\mathrm{Ajetunmobi}}\:\boldsymbol{\mathrm{Abdulqoyyum}} \\ $$

Answered by Ajetunmobi last updated on 16/Feb/21

it should be tetragamma not trigamma sorry for the mistakes

$$\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{should}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{tetragamma}}\:\boldsymbol{\mathrm{not}}\:\boldsymbol{\mathrm{trigamma}} \\ $$$$\boldsymbol{\mathrm{sorry}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{mistakes}} \\ $$

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