# Questions tagged [harmonic-numbers]

For questions regarding harmonic numbers, which are partial sums of the harmonic series. The $N$-th harmonic number is the sum of reciprocals of the first $N$ natural numbers.

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### Can $\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}$, with $H_n$ the $n$-th harmonic number, be written in terms of $\zeta$ values?

The Euler sums are given by $$S_{p,q} = \sum_{n = 1}^{\infty} \frac{H_{n}^{(p)}}{n^q},$$ where $$H_{n}^{(p)} = \sum_{j = 1}^{n} \frac{1}{j^p}.$$ According to Wolfram, Eq. (19), the following special ...
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### On binomial sums $\sum_{n=1}^\infty \frac{1}{n^k\,\binom {2n}n}$ and log sine integrals

Seven years ago, I asked about closed-forms for the binomial sum $$\sum_{n=1}^\infty \frac{1}{n^k\,\binom {2n}n}$$ Some alternative results have been made. Up to a certain $k$, it seems it can be ...
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### Expressions approximating Generalized Harmonic Number (truncated polynomials with shrinking error term preferred)

Specifically, $$H_m^{(2n)} \approx\ ?$$ and $$H_m^{(4n)} \approx\ ?$$ where $(m, n)$ $\in \mathbb N_{>1}$ I would not like to use special functions like the (Riemann zeta function) unless they ...
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### Compute : $\sum_{m≥1}\frac{(-1)^{m}}{(m+1)(2m+1)^{2}}$

How can Compute in closed form this double summation : $\displaystyle\sum_{m≥1}\frac{(-1)^{m}}{(m+1)(2m+1)^{2}}$ I need to evaluate this sum using digamma function Actually I don't have any ideas ...
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### Elementary number th. Harmonic sum and its mod

Let $\frac{1}{2}+\frac{1}{3}+...+\frac{1}{121}=\frac{p}{q}$ where p,q are coprime integer couple. Prove $p\equiv 50 \pmod {121}.$ What im guessing about is that wolstenholme will do its job here but ...
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### Do these polynomials with harmonic number-related coefficients lie in some particular known class?

I've generated a set of univariate polynomials ($b=1,2,\ldots$) in $v$ of degree $b-1$. The constant term and the coefficient of $v^{b-1}$ is simply $H_b$, the $b$-th harmonic number. The ...
I am interested in estimating the following family of sums: $$S_k(n) \equiv \sum_{\substack{n_1, \ldots, n_k \geq 1\\n_1 + \ldots + n_k = n}}\frac{1}{n_1\ldots n_k}$$ where $k \geq 1, n \geq 1$. A ...