Sum of Trigamma of n plus 3 over n plus 2
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Theorem
| \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\map {\psi_1} {n + 3} } {\paren {n + 2} }\) | \(=\) | \(\ds \dfrac {\map {\psi_1} 3} 2 + \dfrac {\map {\psi_1} 4} 3 + \dfrac {\map {\psi_1} 5} 4 + \dfrac {\map {\psi_1} 6} 5 + \cdots\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \zeta 3 - \map \zeta 2 + 1\) |
where:
- $\map {\psi_1} n$ is the trigamma function of $n$
- $\map \zeta 2$ is the Riemann $\zeta$ function of $2$
- $\map \zeta 3$ is Apéry's constant: the Riemann $\zeta$ function of $3$.
Proof
| \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\map {\psi_1} {n + 3} } {\paren {n + 2} }\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \dfrac {\map {\psi_1} {n + 1} } n - \map {\psi_1} 2\) | $n \to n - 2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \dfrac {\map {\psi_1} {n + 1} } n - \paren {\map {\psi_1} 1 - 1}\) | Recurrence Relation for Polygamma Function: $\map {\psi_1} 2 = \map {\psi_1} 1 - 1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \dfrac {\map {\psi_1} {n + 1} } n - \map \zeta 2 + 1\) | Polygamma Function in terms of Hurwitz Zeta Function: $\map {\psi_1} 1 = \map \Gamma 2 \map \zeta {2, 1}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \paren {-\int_0^1 \frac {x^{\paren {n + 1} - 1} \ln x} {n \paren {1 - x} } \rd x} - \map \zeta 2 + 1\) | Integral Form of Polygamma Function: $\ds \map {\psi_1} z = -\int_0^1 \frac {u^{z - 1} \ln u} {1 - u} \rd u$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^1 \paren {-\sum_{n \mathop = 1}^\infty \frac {x^n} n} \frac {\ln x} {\paren {1 - x} } \rd x - \map \zeta 2 + 1\) | Tonelli's Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^1 \frac {\map \ln {1 - x} \ln x} {\paren {1 - x} } \rd x - \map \zeta {2} + 1\) | Power Series Expansion for Logarithm of $1 - x$: $\ds \map \ln {1 - x} = -\sum_{n \mathop = 1}^\infty \frac {x^n} n$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \zeta 3 - \map \zeta 2 + 1\) | Definite Integral: $\ds \int_0^1 \dfrac {\ln x \map \ln {1 - x} } {\paren {1 - x} } \rd x$ |
$\blacksquare$
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