关于无穷级数∑ₙ=0^∞1/(nᵖ+1)的闭形式与zeta函数表示咨询

Great question! Let's break this down for both your specific case ($p=3/2$) and the general non-integer $p$ scenario.

针对$p=3/2$的具体情况 #

First, let's confirm the numerical approximation you mentioned: $\sum_{n=0}^\infty \frac{1}{n^{3/2}+1} \approx 1.95202$. To address your questions:

• Closed-form possibility: Unfortunately, there's no known elementary closed-form for this series. The term $\frac{1}{n^{3/2}+1}$ doesn't simplify to combinations of elementary functions in a way that leads to a finite, closed sum.

• Zeta function representation: This is feasible, though the resulting series is conditionally convergent (its partial sums oscillate before converging to the correct value). Here's how to derive it:

Start by splitting the series into the $n=0$ term and the rest:
$$\sum_{n=0}^\infty \frac{1}{n^{3/2}+1} = 1 + \sum_{n=1}^\infty \frac{1}{n^{3/2}+1}$$

For $n \geq 1$, use the geometric series expansion $\frac{1}{1+x} = \sum_{k=0}^\infty (-1)^k x^k$ (valid for $|x| < 1$, which holds for $n \geq 2$ when $p=3/2$; we handle $n=1$ via Abel summation for conditional convergence):
$$\frac{1}{n^{3/2}+1} = \frac{1}{n^{3/2}} \cdot \frac{1}{1 + n^{-3/2}} = \sum_{k=0}^\infty (-1)^k n^{-3(k+1)/2}$$

Swap the order of summation (justified via Abel summation for conditional convergence):
$$\sum_{n=1}^\infty \frac{1}{n^{3/2}+1} = \sum_{k=0}^\infty (-1)^k \zeta\left(\frac{3(k+1)}{2}\right)$$

Combining with the $n=0$ term, we get:
$$\sum_{n=0}^\infty \frac{1}{n^{3/2}+1} = 1 + \zeta\left(\frac{3}{2}\right) - \zeta(3) + \zeta\left(\frac{9}{2}\right) - \zeta(6) + \zeta\left(\frac{15}{2}\right) - \dots$$

The partial sums of this zeta function series oscillate, but using acceleration techniques (like Euler's transformation) will converge to your numerical approximation of ~1.95202.

一般$p \notin \mathbb{Z}$的通用表示方法 #

For non-integer $p$ with $\text{Re}(p) > 1$ (to ensure the original series converges), we can extend the above approach:

1. Zeta function alternating series #

Similar to the $p=3/2$ case, we split the series and use geometric series expansion, leading to:
$$\sum_{n=0}^\infty \frac{1}{n^p +1} = 1 + \sum_{k=1}^\infty (-1)^{k-1} \zeta(pk)$$

This series is conditionally convergent (since $\zeta(pk) \to 1$ as $k \to \infty$), so you'll need summation acceleration methods to compute its value efficiently. For $\text{Re}(p) \leq 1$, the original series still converges, but we need to use the analytic continuation of the zeta function to handle diverging terms in the alternating series.

2. Integral representation #

Using the integral identity $\frac{1}{x+1} = \int_0^1 t^x dt$, we can rewrite the series as:
$$\sum_{n=0}^\infty \frac{1}{n^p +1} = 1 + \int_0^1 \sum_{n=1}^\infty t^{np} dt$$

The sum $\sum_{n=1}^\infty t^{np}$ is a theta-type function, which doesn't have an elementary closed form for non-integer $p$, but this representation is useful for theoretical analysis or numerical integration.

3. Polylogarithm representation #

For $|r| < 1$, we can write the regularized series (with a convergence factor $r^n$) as:
$$\sum_{n=0}^\infty \frac{r^n}{np +1} = 1 + \sum_{k=0}^\infty (-1)^k \text{Li}_{p(k+1)}(r)$$

Taking the limit as $r \to 1^-$ gives the original series sum (for $\text{Re}(p) > 1$), where $\text{Li}_s(r)$ is the polylogarithm function.

总结 #

• For $p=3/2$, no elementary closed form exists, but you can represent the series as a conditionally convergent alternating sum of zeta functions at half-integer arguments.
• For general non-integer $p$, the series can be expressed via alternating zeta function sums, integral representations involving theta-type functions, or polylogarithm functions (with regularization for smaller $p$).

内容的提问来源于stack exchange，提问作者Franklin Pezzuti Dyer