Question-163435

Question Number 163435 by amin96 last updated on 06/Jan/22

Answered by Mathspace last updated on 07/Jan/22

S=Σ_(n=0) ^∞ (1/(4n+1))((1/(4n+3))−(1/(4n+4))) =Σ_(m=0) ^∞ (1/((4n+1)(4n+3))) −Σ_(n=0) ^∞ (1/((4n+1)(4n+4))) =(1/(16))Σ_(n=0) ^∞ (1/((n+(1/4))(n+(3/4)))) −(1/(16))Σ_(n=0) ^∞ (1/((n+(1/4))(n+1))) we have Σ_(n=0) ^∞ (1/((n+a)(n+b))) =((ψ(a)−ψ(b))/(a−b)) ⇒ Σ_(n=0) ^∞ (1/((n+(1/4))(n+(3/4)))) =((ψ((1/4))−ψ((3/4)))/((1/4)−(3/4)))=(1/2)(ψ((3/4))−ψ((1/4))} Σ_(n=0) ^∞ (1/((n+(1/4))(n+1))) =((ψ((1/4))−ψ(1))/((1/4)−1))=(4/3){ψ(1)−ψ((1/4))} ⇒ S=(1/(32)){ψ((3/4))−ψ((1/4))} −(1/(12)){ψ(1)−ψ((1/4))}

$${S}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{1}}\left(\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{4}}\right) \\ $$$$=\sum_{{m}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{3}\right)} \\ $$$$−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({n}+\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$−\frac{\mathrm{1}}{\mathrm{16}}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({n}+\mathrm{1}\right)} \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+{a}\right)\left({n}+{b}\right)} \\ $$$$=\frac{\psi\left({a}\right)−\psi\left({b}\right)}{{a}−{b}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({n}+\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$=\frac{\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right\} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\psi\left(\mathrm{1}\right)}{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}}=\frac{\mathrm{4}}{\mathrm{3}}\left\{\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right\}\:\Rightarrow \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{32}}\left\{\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right\} \\ $$$$−\frac{\mathrm{1}}{\mathrm{12}}\left\{\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right\} \\ $$$$ \\ $$

Commented by Tawa11 last updated on 07/Jan/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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