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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
46 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 ...
2
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3answers
79 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 ...
-1
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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
57 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; ...
0
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0answers
52 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 ...
32
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3answers
772 views

Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$

One of the possible ways of computing the series is to obtain the generating function, but this might be a tedious, hard work, pretty hard to obtain. What would you propose then? $$\sum_{n=1}^{\...
3
votes
1answer
150 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 ...
3
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2answers
182 views

Computing the integral $\int \limits_{1}^{\infty}\left(\frac{1}{\lfloor{x}\rfloor}-\frac{1}{x}\right)$ …

Prove that $$\large\int \limits_{1}^{\infty}\Bigg(\dfrac{1}{\lfloor{x}\rfloor}-\dfrac{1}{x}\Bigg)dx=\lim \limits_{n \to \infty} \Bigg(-\ln(n) + \sum \limits_{k=1}^n\dfrac{1}{k}\Bigg)$$ I was reading ...
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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 ...
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
149 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 ...
14
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1answer
213 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 ...
4
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1answer
129 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)+...
3
votes
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 ...
2
votes
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 ...
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 \...
10
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3answers
230 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}\...
8
votes
3answers
156 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
vote
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+...+\...
8
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5answers
238 views

How to show $\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}=2$?

I am interested in the proof of $$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}=2, \quad H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$$ This result can be verified by Mathematica or by WolframAlpha This ...
2
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3answers
461 views

Calculating the summation$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}$

I need to find explicitly the following summation $$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}, \quad H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$$ From Mathematica, I checked that the answer is $2$. The ...
0
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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 ...
6
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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\...
3
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5answers
469 views

Closed form for ${\large\int}_0^1\frac{\ln^4(1+x)\ln x}x \, dx$

Can someone compute $$ \int_0^1\frac{\ln^4(1+x)\ln x}x \,dx$$ in closed form? I conjecture that the answer can be expressed as a polynomial function with rational coefficients in constants of the ...
0
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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\...
0
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0answers
94 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)}\...
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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2answers
223 views

Two inequalities for the difference $H_n - H_m$ of harmonic numbers

Let's note the harmonic numbers by $H_n = \sum_{i=1}^n \frac{1}{i}$. Asymptotic expansion of harmonic number is: $(1) H_n = \ln n + \gamma + \frac{1}{2n} - O\left(\frac{1}{n^2}\right)$ Very popular ...
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)^{...
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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 ...
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} =...
3
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2answers
223 views

Proof that $\sum_{k=2}^{\infty} \frac{H_k}{k(k-1)} $ where $H_n$ is the sequence of harmonic numbers converges?

How to prove that $$\displaystyle \sum_{k=2}^{\infty} \dfrac{H_k}{k(k-1)} $$ where $H_n$ is the sequence of harmonic numbers converges and that $\dfrac{H_n}{n(n-1)}\to 0 \ $ I have already proven by ...
14
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2answers
535 views

Prove that ${\sum\limits_{n=1}^{\infty}}(-1)^{n-1} \frac{H_n}{n} = \frac{\pi^2}{12} - \frac12\ln^2 2$

We know that $H_n = \sum_{j=1}^{n}{1 \over j}$. Article in The Sum of Certain Series Related To Harmonic Numbers of Omran Kolba, we have proof of this identity which involves some advanced concepts. ...
1
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1answer
61 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 ...
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 ...
126
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22answers
9k views

Why does the series $\sum_{n=1}^\infty\frac1n$ not converge?

Can someone give a simple explanation as to why the harmonic series $$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$ doesn't converge, on the other hand it grows very slowly?...
13
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2answers
344 views

Generalized Harmonic Number Summation $ \sum_{n=1}^{\infty} {2^{-n}}{(H_{n}^{(2)})^2}$

Prove That $$ \sum_{n=1}^{\infty} \dfrac{(H_{n}^{(2)})^2}{2^n} = \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 $$ Notation : $ \...
4
votes
3answers
130 views

Show that $\sum\limits_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)}=3\zeta(3)$, where, for every positive $n$, $H_n=\sum\limits_{k=1}^n\frac1k$

The problem I was considering about is the evaluation of the following series: \begin{align*} \sum_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)} \end{align*} The attempt I could make was to change $(H_{n})^...
3
votes
4answers
176 views

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 ...
0
votes
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 ...
0
votes
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 ...
3
votes
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}^{\...
25
votes
4answers
1k views

A series involves harmonic number

How do we get a closed form for $$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}$$
1
vote
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
2
votes
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: $$\...
53
votes
5answers
2k views

Infinite Series $\sum\limits_{n=1}^\infty\frac{(H_n)^2}{n^3}$

How to prove that $$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$ $H_n$ denotes the harmonic numbers.
60
votes
12answers
5k views

The limit of truncated sums of harmonic series, $\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}$

What is the sum of the 'second half' of the harmonic series? $$\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?$$ More precisely, what is the limit of the above sequence of partial ...
41
votes
7answers
3k views

Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$

How can I find a closed form for the following sum? $$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$ ($H_n=\sum_{k=1}^n\frac{1}{k}$).
49
votes
7answers
8k views

Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$

I'm trying to find a closed form for the following sum $$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$ where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number. Could you help me with it?