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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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1answer
42 views

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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3answers
71 views

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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0answers
51 views

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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1answer
106 views

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 ...
3
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1answer
149 views

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 ...
2
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1answer
48 views

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 ...
4
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3answers
148 views

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 ...
4
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1answer
128 views

$\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$

This problem was proposed by Cornel and he showed that $$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac13\ln^42-\frac12\ln^22\zeta(2)+\frac72\ln2\zeta(3)-\frac{21}4\zeta(4)+...
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1answer
105 views

Find the closed form of $\quad\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}$

where $H_n$ is the harmonic number and can be defined as: $H_n=1+\frac12+\frac13+...+\frac1n$ $H_n^{(3)}=1+\frac1{2^3}+\frac1{3^3}+...+\frac1{n^3}$ I managed to prove $\quad\displaystyle\sum_{n=1}^...
8
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1answer
182 views

Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$

where $H_n$ is the harmonic number and can be defined as: $H_n=1+\frac12+\frac13+...+\frac1n$ $H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$ these two sums are already solved by Cornel using ...
3
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1answer
198 views

Is there a closed form for a give infinite sum?

I've been asked to evaluate this sum $$\sum_{n=0}^{\infty}\frac{C_n^2}{2^{4n}}(H_{n+1}-H_n)$$ where $C_n=\frac1{n+1}{2n\choose n}$ denotes the $n$th Catalan number and $H_n$ denotes the nth Harmonic ...
8
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3answers
154 views

two challenging sums $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}$

where $H_n$ is the harmonic number and can be defined as: $H_n=1+\frac12+\frac13+...+\frac1n$ $H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$ again, my goal of posting these two challenging ...
1
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1answer
64 views

nice two related sums $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}$

I managed to find the closed forms of these two alternating sums but again my point of posting them is that we can use them as a reference in our solutions. where : $H_n=1+\frac12+\frac13+...+\...
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0answers
48 views

Two Euler type sums involving binomial coefficient and harmonic number

I want to find the closed forms of the following Euler type sums $$\sum\limits_{n=1}^\infty \frac{H_n}{n^p\binom{2n}{n}}4^n$$ and $$\sum\limits_{n=1}^\infty \frac{H_{2n}}{n^p\binom{2n}{n}}4^n.$$ A ...
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2answers
34 views

About the Euler-Mascheroni Constant

Okay, so the limiting difference between the harmonic series and the natural logarithm is known as the Euler-Mascheroni constant, $\gamma= 0,577$. My question is: is there any base for the logarithm ...
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0answers
36 views

A series involving binomial coefficient and harmonic number

I want to find a closed form of the following infinite series $$\sum\limits_{n=1}^\infty \frac{H_{n-1}\binom{2n}{n}}{4^n n^2}=?$$ It can be expressed in terms of $\gamma$ and $\pi$? Here $H_n=\sum\...
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0answers
98 views

Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^5 2^n}$

Given the nth harmonic number $ H_n = \sum_{j=1}^{n} \frac{1}{j}$, we get from this post that apparently, $$\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n= S_{k-1,2}(z) + \rm{Li}_{\,k+1}(z)$$ for $-1\leq z\...
1
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1answer
67 views

Closed form for $f(z,a)=\sum_{n=1}^\infty \frac{H_n}{n^a}z^n$?

So I have been working with some power series involving the Harmonic numbers. I have been able to evaluate the first couple of sums as $$f(z,0)=\sum_{n=1}^\infty H_n z^n=\sum_{n=1}^\infty z^n\int_0^1 \...
5
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1answer
110 views

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 ...
2
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1answer
40 views

Does the divergence of one partial sum imply overall divergence?

Consider a sequence defined by $$a_{3n-2}=-\frac{1}{2n-1}, a_{3n-1} = \frac{1}{4n-1}, a_{3n} = \frac{1}{4n}$$ I want to show $\sum a_n$ diverges. Is it enough the show that the partial sum $S_{3k} =...
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0answers
93 views

prove $\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$

this series was evaluated by Cornel Valean here using series manipulation. I took a different path as follows: using the identity:$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\...
1
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1answer
60 views

prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$

I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions. other approaches ...
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0answers
118 views

Find limit of sequence defined by sum of previous terms and harmonics

I came across this sequence as part of my work. Could someone indicate me the methodology I should follow to solve it? I guess it involves harmonic numbers and/or the digamma function? I tried to ...
3
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1answer
107 views

Closed form for sum of powers

I wonder if it is possible to evaluate explicitly the sum $$S(N):=\sum_{j=1}^{\left\lfloor\frac{N-1}{2}\right\rfloor}\left(1-\frac{2j}{N}\right)^{N+1},\quad N\in\mathbb{N}.$$ In the large $N$ limit ...
0
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1answer
69 views

Double harmonic series $\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{H_{n+m}^{(p)}}{(n+1)^{q}(m+1)^{r}}$

Do these sums exist in the literature and have been investigated before? The same question for the odd variant, that is $$ \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{O_{n+m}^{(p)}}{(2n+1)^{q}(2m+1)^{...
3
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1answer
56 views

A “binomial” generalization of harmonic numbers

For positive integers $s$ and $n$ (let's limit the generality), define $$H_s(n)=\sum_{k=1}^{n}\frac{1}{k^s},\qquad G_s(n)=\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k-1}}{k^s}.$$ The former is well-known; ...
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0answers
44 views

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 ...
3
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2answers
59 views

Show that : $\sum_{k=1}^{\infty}\frac{i^{k(5k+1)}}{k(k+1)}=1-\frac{π}{2}$

Show that $S=\displaystyle\sum_{k=1}^{\infty}\frac{i^{k(5k+1)}}{k(k+1)}=1-\frac{π}{2}$ My try : $S=\displaystyle\sum_{k=1}^{\infty}\frac{e^{iπk(5k+1)/2}}{k(k+1)}$ $=\displaystyle\sum_{k=1}^{\...
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2answers
78 views

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 ...
1
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1answer
61 views

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
14
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1answer
212 views

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 ...
2
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3answers
124 views

Compute in closed form $\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$

I am trying to find closed form for this integral: $$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$ Where $a>0$. My try: Let: $$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$ Then: $$\...
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2answers
22 views

Is there a further simplification for alternating harmonics with order?

I know that the harmonic number $H_a ^{(b)}$ is $$\sum_{n=1}^a \frac{1}{n^b}$$ I was wondering if, for the generalized alternating harmonic number $\bar H_a^{(b)}$, there was a closed formula. For ...
0
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2answers
28 views

Asymptotic analysis of harmonic series using Calculus

The problem is to proof that Harmonic series $\sum_{i=1}^n \frac{1}{i} = O(ln \space n)$ So, I know that $ln \space n = \int_{1}^n \frac{1}{x} dx$ so, I need to prove that $H(n) = 1+\frac{1}{2}+...+...
3
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0answers
53 views

How to evaluate $\sum_{n=1}^{\infty}\frac{H_{kn}^2-[\gamma+\ln(kn)]^2}{n}?$

From this post @Olivier Oloa gives the closed form for this sum $(1)$ $$\sum_{n=1}^{\infty}\frac{H_n^2-(\gamma+\ln n)^2}{n}=\frac{5}{3}\zeta(3)-\frac{2}{3}\gamma^3-2\gamma\gamma_1-\gamma_2\tag1$$ I ...
10
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3answers
229 views

Evaluating $\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}$

Question: How can we evaluate $$\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n},$$where $H_n=\frac11+\frac12+\cdots+\frac1n$? Quick Results This series converges because $$\frac{(H_n)^2}{n}\...
0
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0answers
28 views

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 ...
4
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2answers
151 views

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 ...
2
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0answers
97 views

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 $$ H_{p-1}:= \sum_{i=1}^{p-1}\frac{1}{i}\equiv 0 \pmod {p^2}$$ ...
2
votes
3answers
68 views

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 ...
2
votes
1answer
84 views

How to Prove Integral Representations of $H_z$ and $\psi(z)$

I've run across several integral representations of $H_z$ such as the following (see Harmonic Number Integral Representations). (1) $\quad H_z=\int\limits_0^1 \frac{1-t^z}{1-t} \, dt\,,\qquad\qquad\...
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1answer
35 views

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 ...
1
vote
0answers
30 views

Leave-one-out Harmonic number

I am trying to find a number sequence $x_k\geq 0$ such that $$\text{for all integers }n\geq k\geq 1,\quad H_{n-1 - x_k(n-k)}\geq H_n-\frac{1}{k}.$$ I recall that the harmonic numbers $s\mapsto H_s$ ...
3
votes
0answers
88 views

Evaluating $\int_{0}^{1-x} \frac{1}{1-x_{n}} \cdots \int_{0}^{1-x_3} \frac{1}{1-x_2} \int_{0}^{1-x_2} \frac{1}{1-x_1} dx_1 dx_2 \cdots dx_n$

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<...
1
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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 ...
1
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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 ...
0
votes
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 ...
0
votes
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 ...
8
votes
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) \...
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votes
1answer
69 views

Is there a formula for a summation divided by a product of its terms?

$$\frac{\sum_{i=1}^{n}x_{i}}{\prod_{i=1}^{n}x_{i}}= \frac{1}{x_{2}x_{3}x_{4}...}+\frac{1}{x_{1}x_{3}x_{4}}+\frac{1}{x_{1}x_{2}x_{4}}...$$ There is a very clear pattern that each consecutive result ...