Menu Close

is-there-a-formula-for-this-n-0-1-n-2-an-b-

Tinku Tara 0 Comments
Pin




Question Number 122861 by MJS_new last updated on 20/Nov/20
is there a formula for this? Σ_(n=0) ^∞ (1/(n^2 +an+b))
$$\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{this}? \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{an}+{b}} \\ $$
Commented by Dwaipayan Shikari last updated on 20/Nov/20
n^2 +an+b=(n+((a−(√(a^2 −4b)))/2))(n+((a+(√(a^2 −4b)))/2))=(n+K_1 )(n+K_2 ) (1/(K_2 −K_1 ))Σ_(n=0) ^∞ (1/(n+K_1 ))−(1/(n+K_2 ))=(1/(K_1 −K_2 ))Σ_(n=1) ^∞ (1/(n−1+K_1 ))−(1/(K_1 −K_2 ))Σ^∞ (1/(n−1+K_2 )) ((1/(K_1 −K_2 ))Σ^∞ (1/n)−(1/(n−1+K_2 )))−(1/(K_1 −K_2 ))(Σ^∞ (1/n)−(1/(n−1+K_1 ))) (1/(K_1 −K_2 ))(ψ(K_2 −1)−ψ(K_1 −1)) (1/( (√(a^2 −4b))))(ψ((a/2)−((√(a^2 −4b))/2)−1)−ψ((a/2)+((√(a^2 −4b))/2)−1))
$${n}^{\mathrm{2}} +{an}+{b}=\left({n}+\frac{{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}}\right)\left({n}+\frac{{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}}\right)=\left({n}+{K}_{\mathrm{1}} \right)\left({n}+{K}_{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{1}}{{K}_{\mathrm{2}} −{K}_{\mathrm{1}} }\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+{K}_{\mathrm{1}} }−\frac{\mathrm{1}}{{n}+{K}_{\mathrm{2}} }=\frac{\mathrm{1}}{{K}_{\mathrm{1}} −{K}_{\mathrm{2}} }\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\mathrm{1}+{K}_{\mathrm{1}} }−\frac{\mathrm{1}}{{K}_{\mathrm{1}} −{K}_{\mathrm{2}} }\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}−\mathrm{1}+{K}_{\mathrm{2}} } \\ $$$$ \\ $$$$\left(\frac{\mathrm{1}}{{K}_{\mathrm{1}} −{K}_{\mathrm{2}} }\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}−\mathrm{1}+{K}_{\mathrm{2}} }\right)−\frac{\mathrm{1}}{{K}_{\mathrm{1}} −{K}_{\mathrm{2}} }\left(\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}−\mathrm{1}+{K}_{\mathrm{1}} }\right) \\ $$$$ \\ $$$$\:\:\frac{\mathrm{1}}{{K}_{\mathrm{1}} −{K}_{\mathrm{2}} }\left(\psi\left({K}_{\mathrm{2}} −\mathrm{1}\right)−\psi\left({K}_{\mathrm{1}} −\mathrm{1}\right)\right) \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}\left(\psi\left(\frac{{a}}{\mathrm{2}}−\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}}−\mathrm{1}\right)−\psi\left(\frac{{a}}{\mathrm{2}}+\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}}−\mathrm{1}\right)\right) \\ $$
Commented by MJS_new last updated on 20/Nov/20
thank you so much!
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}! \\ $$
Commented by Dwaipayan Shikari last updated on 20/Nov/20
�� sir!

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *