What would be the value of ∑n=0∞1an2+bn+c

Question

Rohan Mercado · Accepted Answer

f(z)=1az2+bz+c  The poles of f(z) are located at  z0=−b+b2−4ac2a  and z1=−b−b2−4ac2a  Then b0=Resz=z0πcot⁡(πz)f(z)=limz→z0z−z0πcot⁡(πz)az2+bz+c=limz→z0πcot⁡(πz)+(z0−z)π2csc2(πz)2az+b  Using LHopitals rule. Continuing, we have the limit is  limz→z0πcot⁡(πz)+(z0−z)π2csc2(πz)2az+b=πcot⁡(πz0)2z0+b  For z0≠0  Similarly, we find  b1=Resz=z1πcot⁡(πz)f(z)=πcot⁡(πz1)2az1+b  Then ∑n=−∞∞1an2+bn+c=−(b0+b1)=−π(cot⁡(πz0)2z0+b+cot⁡(πz1)2az1+b)=−π(cot⁡(πz0)b2−4ac+cot⁡(πz1)−b2−4ac)

trasahed · Accepted Answer

The solution that follows considers the sum

\sum_{n = - \infty}^{\infty} \frac{1}{a n^{2} + b n + c}

, and throughout I will write

\sum_{n = - \infty}^{\infty} f (n)

to mean

lim_{N \to \infty} \sum_{n = - N}^{N} f (n)

.
Factoring the quadratic, with your definition of

z_{0}, z_{1}

, we have

\sum_{n = - \infty}^{\infty} \frac{1}{a n^{2} + b n + c} = \frac{1}{a} \sum_{n = - \infty}^{\infty} \frac{1}{(n - z_{0}) (n - z_{1})}

Assume that neither

z_{0} nor z_{1}

are integers, since otherwise we would have a

\frac{1}{0}

term appearing in the sum. By applying partial fractions, remembering that

z_{0} - z_{1} = \frac{\sqrt{b^{2} - 4 a c}}{a}

we get

\frac{1}{\sqrt{b^{2} - 4 a c}} \sum_{n = - \infty}^{\infty} (\frac{1}{n - z_{0}} - \frac{1}{n - z_{1}})

By the cotangent identity

π \cot (π x) = \sum_{n = - \infty}^{\infty} \frac{1}{n + x}

, we conclude that

\sum_{n = - \infty}^{\infty} \frac{1}{a n^{2} + b n + c} = \frac{π \cot (π z_{1}) - π \cot (π z_{0})}{\sqrt{b^{2} - 4 a c}}

RizerMix · Accepted Answer

Take a=1,b=3,c=2,then z0=−2,z1=−1, and so you have to compute cot⁡(−π) and cot⁡(−2π) which make no sence, However ∑n=0∞1n2+3n+2=∑n=0∞(1n+1−1n+2)=limm→∞(1−1m+2)=1

What would be the value of \sum_{n=0}^{\infty} \frac{1}{an^{2}+bn+c}

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