Finding ∑n=1∞1f(n) where f is a real quadratic function?

an2+bn+c=a(n−r)(n−s)where r and s are the roots of the quadratic (hoping that they are not psitive integer numbers) and use partial fraction decomposition1an2+bn+c=1a(r−s)(1n−r−1n−s)So, for the partial sumSp=∑n=1p1an2+bn+c=1a(r−s)(ψ(p−r+1)−ψ(1−r)−ψ(p−s+1)+ψ(1−s))Now, using the asymptoticsSp=1a(r−s)((ψ(1−s)−ψ(1−r))+s−rp+O(1p2))S∞=ψ(1−s)−ψ(1−r)a(r−s)Back to a, b, cS∞=ψ(b+b2−4ac2a+1)ψ(b−b2−4ac2a+1)b2−4acor S∞=Hb+b2−4ac2a−Hb−b2−4ac2ab2−4ac

If&nbsp;a=0, the series diverges. If&nbsp;a≠0, the sum can be expressed using the values of the digamma function&nbsp;(γ(0,...))&nbsp;see Wolfram alpha:∑n=1m1an2+bn+c=d⋅γ(0,b2a+m−d2a+1)−d⋅γ(0,b2a+m+d2a+1)−d⋅γ(0,b2a−d2a+1)+d⋅γ(0,b2a+d2a+1)d2where&nbsp;d=b2−4ac.

an2+bn+c=a(n−r)(n−s)where r and s are the roots of the quadratic (hoping that they are not psitive integer numbers) and use partial fraction decomposition1an2+bn+c=1a(r−s)(1n−r−1n−s)So, for the partial sumSp=∑n=1p1an2+bn+c=1a(r−s)(ψ(p−r+1)−ψ(1−r)−ψ(p−s+1)+ψ(1−s))Now, using the asymptoticsSp=1a(r−s)((ψ(1−s)−ψ(1−r))+s−rp+O(1p2))S∞=ψ(1−s)−ψ(1−r)a(r−s)Back to a, b, cS∞=ψ(b+b2−4ac2a+1)ψ(b−b2−4ac2a+1)b2−4acor S∞=Hb+b2−4ac2a−Hb−b2−4ac2ab2−4ac

If&nbsp;a=0, the series diverges. If&nbsp;a≠0, the sum can be expressed using the values of the digamma function&nbsp;(γ(0,...))&nbsp;see Wolfram alpha:∑n=1m1an2+bn+c=d⋅γ(0,b2a+m−d2a+1)−d⋅γ(0,b2a+m+d2a+1)−d⋅γ(0,b2a−d2a+1)+d⋅γ(0,b2a+d2a+1)d2where&nbsp;d=b2−4ac.

Find out the answer to "Finding ∑n=1∞1f(n) where f is a real quadratic..."

High School
Algebra
Algebra I
Quadratic function and equation

Keenan Rhodes

Answered question

2022-04-05

Finding

\sum_{n = 1}^{\infty} \frac{1}{f (n)}

where f is a real quadratic function?

Answer & Explanation

libertydragonrbha

Beginner2022-04-06Added 15 answers

a n^{2} + b n + c = a (n - r) (n - s)

where r and s are the roots of the quadratic (hoping that they are not psitive integer numbers) and use partial fraction decomposition

\frac{1}{a n^{2} + b n + c} = \frac{1}{a (r - s)} (\frac{1}{n - r} - \frac{1}{n - s})

So, for the partial sum

S_{p} = \sum_{n = 1}^{p} \frac{1}{a n^{2} + b n + c} = \frac{1}{a (r - s)} (ψ (p - r + 1) - ψ (1 - r) - ψ (p - s + 1) + ψ (1 - s))

Now, using the asymptotics

S_{p} = \frac{1}{a (r - s)} ((ψ (1 - s) - ψ (1 - r)) + \frac{s - r}{p} + O (\frac{1}{p^{2}}))

S_{\infty} = \frac{ψ (1 - s) - ψ (1 - r)}{a (r - s)}

Back to a, b, c

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

S_{\infty} = \frac{H_{\frac{b + \sqrt{b^{2} - 4 a c}}{2 a}} - H_{\frac{b - \sqrt{b^{2} - 4 a c}}{2 a}}}{\sqrt{b^{2} - 4 a c}}

Ferrito90gn

Beginner2022-04-07Added 12 answers

If $a = 0$ , the series diverges. If $a \neq 0$ , the sum can be expressed using the values of the digamma function $(γ (0, . . .))$ see Wolfram alpha:
$\sum_{n = 1}^{m} \frac{1}{a n^{2} + b n + c} = \frac{d \cdot γ (0, \frac{b}{2 a} + m - \frac{d}{2 a} + 1) - d \cdot γ (0, \frac{b}{2 a} + m + \frac{d}{2 a} + 1) - d \cdot γ (0, \frac{b}{2 a} - \frac{d}{2 a} + 1) + d \cdot γ (0, \frac{b}{2 a} + \frac{d}{2 a} + 1)}{d^{2}}$
where $d = \sqrt{b^{2} - 4 a c}$ .

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