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

Question Number 122861 by MJS_new last updated on 20/Nov/20

is there a formula for this ?

\underset{n = 0}{\sum^{\infty}} \frac{1}{n^{2} + a n + b}

Commented by Dwaipayan Shikari last updated on 20/Nov/20

n^{2} + a n + b = (n + \frac{a - \sqrt{a^{2} - 4 b}}{2}) (n + \frac{a + \sqrt{a^{2} - 4 b}}{2}) = (n + K_{1}) (n + K_{2})

\frac{1}{K_{2} - K_{1}} \underset{n = 0}{\sum^{\infty}} \frac{1}{n + K_{1}} - \frac{1}{n + K_{2}} = \frac{1}{K_{1} - K_{2}} \underset{n = 1}{\sum^{\infty}} \frac{1}{n - 1 + K_{1}} - \frac{1}{K_{1} - K_{2}} \sum^{\infty} \frac{1}{n - 1 + K_{2}}

(\frac{1}{K_{1} - K_{2}} \sum^{\infty} \frac{1}{n} - \frac{1}{n - 1 + K_{2}}) - \frac{1}{K_{1} - K_{2}} (\sum^{\infty} \frac{1}{n} - \frac{1}{n - 1 + K_{1}})

\frac{1}{K_{1} - K_{2}} (ψ (K_{2} - 1) - ψ (K_{1} - 1))

\frac{1}{\sqrt{a^{2} - 4 b}} (ψ (\frac{a}{2} - \frac{\sqrt{a^{2} - 4 b}}{2} - 1) - ψ (\frac{a}{2} + \frac{\sqrt{a^{2} - 4 b}}{2} - 1))

Commented by MJS_new last updated on 20/Nov/20

thank you so much!

Commented by Dwaipayan Shikari last updated on 20/Nov/20

�� sir!