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.

582 questions
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Finding $f(x)$ that makes the limit $\lim_{x \to \infty} g(x) = e$ and Satisfies Other Condtions

So, for this proof I am working on, I encountered this equation: $$g(x)=(e^{H(x)} \cdot H(x)^{({e^{H(x)}})})^{f(x)}$$ where $H(x)$ is the harmonic series-- $\sum_{n=1}^{x}\dfrac{1}{n}$. So, for me to ...
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Prove $\frac{\partial}{\partial m}\text{B}(n,m)=-\text{B}(n,m)\sum_{k=0}^{n-1}\frac{1}{k+m}$

where $\ \displaystyle\text{B}(n,m)=\int_0^1 x^{n-1}(1-x)^{m-1}\ dx=\frac{\Gamma(n)\Gamma(m)}{\Gamma(n+m)}\$is the beta function, defined over positive $\ n,m>0$. The point of this post is to ...
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evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$

where $H_n=1+\frac1{2}+\frac1{3}+...+\frac1{n}$ is the $n$th harmonic number. this sum was proposed by Cornel and I solved it using integration. can be solved using series manipulation? here is the ...
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Finding $f(x)$ that makes the limit $\lim_{x \to \infty} g(x) = e$ [on hold]

So, with this proof I am working on, I have this equation: $$g(x)=(e^{H(x)}*H(x)^{({e^{H(x)}})})^{f(x)}$$ where $H(x)$ is the harmonic series-- $\sum_{n=1}^{x}\dfrac{1}{n}$. So, for me to continue, I ...
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challenging sum $\sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}$

where $H_n^{(m)}=1+\frac1{2^m}+\frac1{3^m}+...+\frac1{n^m}$ is the $n$th harmonic number of order $m$. this problem was proposed by Cornel Valean on his FB page. I tried to solve it using ...
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A double sum involving harmonic and zeta function

I got this sum from a site, but I can't remember it. $$\sum_{j=1}^{\infty}\frac{1}{2j+1}\sum_{i=1}^{\infty}(-1)^i[H_{i,2j}-\zeta(2j)]=\frac{1-\ln(2)}{2}$$ This looking interesting, but is this sum ...
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Sum $\sum\limits_{n=1}^\infty\frac{H_n^2}{n^22^n}$

Where $H_n$ is the harmonic number, $\ \displaystyle H_n=1+\frac12+\frac13+...+\frac1n$. I am going to present my solution as I need it as a reference. Other approaches are appreciated. here is ...
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On generalizing the harmonic sum $\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n = S_{k-1,2}(1)+\zeta(k+1)$ when $z=1$?

Given the nth harmonic number $H_n = \sum_{j=1}^{n} \frac{1}{j}$. In this post it asks for the evaluation, $$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\tfrac{5}{4}\zeta(4)$$ while this post and this ...
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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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Compute $\sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}$

How can Compute in closed form this double summation : $$\sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}$$ I think here can use harmonic series Actually I don't have any ideas to approach it
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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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On the limit of $H_n^p\over n^q$ and $H_{n^p}\over n^q$

As you know, $H_n$ is the famous Harmonic number defined as follows:$$H_n=\sum_{k=1}^{n}{1\over k}$$I was wondering for which $p,q\in \Bbb N$ do the following limits exist and are equal to some ...
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Evaluate $\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$

By accident, I find this summation when I pursue the particular value of $-\operatorname{Li_2}(\tfrac1{2})$, which equals to integral $\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x}$. Notice this ...
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A sort of Wolstenholme's $p^2$-congruence.

It is well-known (Wolstenholme Theorem) that for any prime $p$ such that $p>3$ the Harmonic number $H_{p-1}$ satisfies the congruence $$H_{p-1}:= \sum_{i=1}^{p-1}\frac{1}{i}\equiv 0 \pmod {p^2}$$ ...
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How can i prove this $\lim\limits_{n\to \infty}\dfrac{\sum_{k=1}^{n} \left(\frac1k\right)}{\sum_{k=1}^{n}{\sin \left(\frac1k\right)}}=1$?

I have accrossed the following sum in my textbook $\lim\limits_{n\to \infty}\dfrac{\sum_{k=1}^{n} \left(\frac1k\right)}{\sum_{k=1}^{n}{\sin \left(\frac1k\right)}}=1$ , I have tried to evaluate ...
I am trying to evaluate the following integral $$\int_{0}^{1-x} \frac{1}{1-x_{n}} \cdots \int_{0}^{1-x_3} \frac{1}{1-x_2} \int_{0}^{1-x_2} \frac{1}{1-x_1} dx_1 dx_2 \cdots dx_{n}, \hspace{0.5cm} 0<... 2answers 90 views Do these polynomials with harmonic number-related coefficients lie in some particular known class? I've generated a set of univariate polynomials (b=1,2,\ldots) in v of degree b-1. The constant term and the coefficient of v^{b-1} is simply H_b, the b-th harmonic number. The ... 1answer 49 views Estimate for multiple harmonic sum I am interested in estimating the following family of sums:$$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}$$where k \geq 1, n \geq 1. A ... 0answers 32 views Harmonic series (double) I am wondering about the \Theta class (i.e. asymptotic complexity) of the following:$$\sum^n_{i=1} \sum_{j=1}^n \frac{1}{ij}$$Since this is basically the harmonic series, applied twice, it ... 0answers 28 views Harmonic numbers Is there a formula for calculating such a sequence of numbers: 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... 1/x? I know that the sum of the Harmonic series is equal to infinity, but is there a formula for ... 1answer 123 views Two Euler sums each containing the reciprocal of the central binomial coefficient Is it possible to find closed-form expressions for the following two Euler sums containing the reciprocal of the central binomial coefficient?$$1. \sum_{n = 0}^\infty \frac{(-1)^n H_n}{(2n + 1) \...
$$\frac{\sum_{i=1}^{n}x_{i}}{\prod_{i=1}^{n}x_{i}}= \frac{1}{x_{2}x_{3}x_{4}...}+\frac{1}{x_{1}x_{3}x_{4}}+\frac{1}{x_{1}x_{2}x_{4}}...$$ There is a very clear pattern that each consecutive result ...