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Consider:

$$h(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$$

Find the coefficients $a_n$ of the Laurent Series of $h(z)$ centered at $z=-2$

I got this from the approach here: Infinite sum complex analysis

You see the complex-analytic approach. My question is, why does @MarkoRiedel

Only use $\psi(-z)$ part of the function $h(z)$ in finding the coeffcients. As you see, his coefficient integral is:

$$\frac{1}{2\pi i} \cdot \int_{|z+2| = \epsilon} \frac{\psi(-z)}{(z+2)^{n+1}} dz $$

He does not consider $$\frac{1}{(z+1)(z+2)^3} \space \text{of} \space h(z) \space \text{just the} \space \psi(-z) \space \text{digamma.}$$

How and why?

Thanks!

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Amad27
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  • Marko Riedel does not compute the coefficients with that integral only. He explicitly states that the coefficient is given by computing the coefficients for $\frac{\psi(-z)}{(z+2)^3}$ and the coefficients for $\frac{1}{z+1}=-\frac{1}{1-(z+2)}$. – Jack D'Aurizio Jan 13 '15 at 16:01
  • But why seperate the whole integral then? – Amad27 Jan 13 '15 at 16:02
  • Because computations are easier that way. – Jack D'Aurizio Jan 13 '15 at 16:03
  • @JackD'Aurizio, consider this.

    $$h(z) = ... + a_{-1}(z+2)^{-1} + a_0 + ....$$

    Multiply by $z+2$ to get:

    $$\frac{\psi(-z)}{(z+1)(z+2)^2} =... + a_{-1} + a_0(z+2) + ....$$

    $$\lim_{z \to -2} \frac{\psi(-z)}{(z+1)(z+2)^2} = a_{-1}$$

    If only I could figure out this limit...

     – Amad27 Jan 13 '15 at 16:10
  • I apologize for not having seen this sooner. I would say learn about the so-called Cauchy product which is a very simple concept yet part of the syllabus here in Germany. If you write down the series for $\psi(-z)/(z+2)^3$ and $1/(1+z)$ then the Cauchy product will let you read off the coefficient on $(z+2)^{-1}.$ This is also at Wikipedia. – Marko Riedel Jan 13 '15 at 20:24
  • @MarkoRiedel: that is exactly what I did in my answer. – robjohn Jan 13 '15 at 21:07

1 Answers1

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We can write the series for $\psi(-z)$ at $z=-2$ as $$ \begin{align} \psi(-z) &=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+2-(z+2)}\right)\\ &=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}+\frac1{(z+2)-(k+2)}\right)\\ &=1-\gamma+\sum_{k=1}^\infty(1-\zeta(k+1))(z+2)^k\tag{1} \end{align} $$ where we have used that $$ \begin{align} \frac1{n!}\frac{\mathrm{d}^n}{\mathrm{d}z^n}\sum_{k=0}^\infty\frac1{(z+2)-(k+2)} &\stackrel{\hphantom{z=-2}}{=}\sum_{k=0}^\infty\frac{(-1)^n}{((z+2)-(k+2))^{n+1}}\\ &\stackrel{z=-2}{=}1-\zeta(n+1)\tag{2} \end{align} $$

Therefore, we get $$ \begin{align} \frac{\psi(-z)}{((z+2)-1)(z+2)^3} &=-\left[\sum_{k=0}^\infty(z+2)^{k-3}\right]\left[1-\gamma+\sum_{k=1}^\infty(1-\zeta(k+1))(z+2)^k\right]\\ &=\bbox[5px,border:2px solid #C00000]{\sum_{n=0}^\infty\left[-(1-\gamma)+\sum_{k=1}^n(\zeta(k+1)-1)\right](z+2)^{n-3}}\tag{3} \end{align} $$

I see that this result is used to compute a sum that is far more simply computed using partial fractions. $$ \begin{align} \sum_{n=1}^\infty\frac1{(n+1)(n+2)^3} &=\sum_{n=1}^\infty\left(\color{#C00000}{\frac1{n+1}-\frac1{n+2}}\color{#00A000}{-\frac1{(n+2)^2}}\color{#0000FF}{-\frac1{(n+2)^3}}\right)\\ &=\color{#C00000}{\frac12}\color{#00A000}{-\left(\zeta(2)-1-\frac14\right)}\color{#0000FF}{-\left(\zeta(3)-1-\frac18\right)}\\ &=\frac{23}8-\frac{\pi^2}6-\zeta(3)\tag{4} \end{align} $$

robjohn
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