Skip to main content
added 84 characters in body
Source Link
David E Speyer
  • 156.4k
  • 14
  • 422
  • 763

As GH from MO says, it depends what you think is a closed form.

Any function of the form $P(n)/Q(n)$ with $\deg P \leq \deg Q -2$ can be written as a linear combination of $1/(n+1) - 1/(n+\alpha)$ and $1/(n+\alpha)^k$, for various $\alpha$ and various $k>1$. (useUse partial fraction decomposition). The bound on $\deg P$ ensures that the coefficients of $1/(n+1)$ will cancel out.)

The sum $\sum \left( \frac{1}{n+1} - \frac{1}{n+\alpha} \right)$ is $\gamma+\psi(\alpha)$, where $\psi$ is the digamma function. The sum $\sum \frac{1}{(n+\alpha)^k}$ is the polygamma function $\psi^k(z)$. In both cases, it is worth noting the recurrence identities and reflection relations in the linked Wikipedia articles, which let you relate values at $\alpha$ to values at $m \pm \alpha$, for $m$ an integer.

As GH from MO says, it depends what you think is a closed form.

Any function of the form $P(n)/Q(n)$ with $\deg P \leq \deg Q -2$ can be written as a linear combination of $1/(n+1) - 1/(n+\alpha)$ and $1/(n+\alpha)^k$, for various $\alpha$ and various $k>1$ (use partial fraction decomposition).

The sum $\sum \left( \frac{1}{n+1} - \frac{1}{n+\alpha} \right)$ is $\gamma+\psi(\alpha)$, where $\psi$ is the digamma function. The sum $\sum \frac{1}{(n+\alpha)^k}$ is the polygamma function $\psi^k(z)$. In both cases, it is worth noting the recurrence identities and reflection relations in the linked Wikipedia articles, which let you relate values at $\alpha$ to values at $m \pm \alpha$, for $m$ an integer.

As GH from MO says, it depends what you think is a closed form.

Any function of the form $P(n)/Q(n)$ with $\deg P \leq \deg Q -2$ can be written as a linear combination of $1/(n+1) - 1/(n+\alpha)$ and $1/(n+\alpha)^k$, for various $\alpha$ and various $k>1$. (Use partial fraction decomposition. The bound on $\deg P$ ensures that the coefficients of $1/(n+1)$ will cancel out.)

The sum $\sum \left( \frac{1}{n+1} - \frac{1}{n+\alpha} \right)$ is $\gamma+\psi(\alpha)$, where $\psi$ is the digamma function. The sum $\sum \frac{1}{(n+\alpha)^k}$ is the polygamma function $\psi^k(z)$. In both cases, it is worth noting the recurrence identities and reflection relations in the linked Wikipedia articles, which let you relate values at $\alpha$ to values at $m \pm \alpha$, for $m$ an integer.

Source Link
David E Speyer
  • 156.4k
  • 14
  • 422
  • 763

As GH from MO says, it depends what you think is a closed form.

Any function of the form $P(n)/Q(n)$ with $\deg P \leq \deg Q -2$ can be written as a linear combination of $1/(n+1) - 1/(n+\alpha)$ and $1/(n+\alpha)^k$, for various $\alpha$ and various $k>1$ (use partial fraction decomposition).

The sum $\sum \left( \frac{1}{n+1} - \frac{1}{n+\alpha} \right)$ is $\gamma+\psi(\alpha)$, where $\psi$ is the digamma function. The sum $\sum \frac{1}{(n+\alpha)^k}$ is the polygamma function $\psi^k(z)$. In both cases, it is worth noting the recurrence identities and reflection relations in the linked Wikipedia articles, which let you relate values at $\alpha$ to values at $m \pm \alpha$, for $m$ an integer.