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.

181
votes
10answers
27k views

Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?

If $n>1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer. If you know Bertrand's Postulate, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ ...
129
votes
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?...
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 ...
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.
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?
48
votes
2answers
1k views

Conjectured value of a harmonic sum $\sum_{n=1}^\infty\left(H_n-\,2H_{2n}+H_{4n}\right)^2$

There is a known asymptotic expansion of harmonic numbers $H_n$ for $n\to\infty$: $$\begin{align}H_n&=\gamma+\ln n+\sum_{k=1}^\infty\left(-\frac{B_k}{k\cdot n^k}\right)\\ &=\gamma+\ln n+\frac1{...
43
votes
5answers
1k views

A series involving the harmonic numbers : $\sum_{n=1}^{\infty}\frac{H_n}{n^3}$

Let $H_{n}$ be the nth harmonic number defined by $ H_{n} := \sum_{k=1}^{n} \frac{1}{k}$. How would you prove that $$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}?$$ Simply replacing $H_{n}$ ...
43
votes
5answers
8k views

Do harmonic numbers have a “closed-form” expression?

One of the joys of high-school mathematics is summing a complicated series to get a “closed-form” expression. And of course many of us have tried summing the harmonic series $H_n =\sum \limits_{k \leq ...
43
votes
2answers
1k views

there exist infinite many $n\in\mathbb{N}$ such that $S_n-[S_n]<\frac{1}{n^2}$

recent conjecture :Let $S_n=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$, where $n$ is a positive integer. Prove that :there exist infinite many $n\in\mathbb{N^{+}}$ such that $$S_n-[S_n]<\dfrac{1}...
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? $$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$ ($H_n=\sum_{k=1}^n\frac{1}{k}$).
33
votes
5answers
11k views

How to find the sum of this series : $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$

Problem : How to find the sum of this series : $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$ This is a Harmonic progression : So is this formula correct to sum the series : $\frac{(...
32
votes
4answers
835 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}^{\...
30
votes
4answers
1k views

Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$

How can I prove that $$\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2).$$ Can anyone help me please?
29
votes
6answers
888 views

What would Gauss do in this case: adding $1+\frac12+\frac13+\frac14+ \dots +\frac1{100}$?

We all know the story related to Gauss that Gauss' class was asked to find the sum of the numbers from $1$ to $100$ as a "busy work" problem and and he came up with $5050$ in less than a minute. He ...
28
votes
1answer
2k views

Sum of Squares of Harmonic Numbers

Let $H_n$ be the $n^{th}$ harmonic number, $$ H_n = \sum_{i=1}^{n} \frac{1}{i} $$ Question: Calculate the following $$\sum_{j=1}^{n} H_j^2.$$ I have attempted a generating function approach but ...
28
votes
6answers
1k views

A series involves harmonic number

How do we get a closed form for $$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}$$
27
votes
2answers
844 views

How find this series $\sum_{n=1}^{\infty}\frac{1}{n^2H_{n}}$?

Question: This follow series have simple closed form? $$\sum_{n=1}^{\infty}\dfrac{1}{n^2H_{n}}$$ where $$H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$ I suddenly thought of this ...
25
votes
2answers
2k views

Showing that $\lim_{n\to\infty}\sum^n_{k=1}\frac{1}{k}-\ln(n)=0.5772\ldots$

How to show that $$\lim_{n\to\infty}\left[\sum^n_{k=1}\frac{1}{k}-\ln(n)\right]=0.5772\ldots$$ No clue at all. Need help! Appreciated!
24
votes
2answers
716 views

Proving that $~\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{k^2}~H_k~H_{k-1}=\frac{3}{16}~\zeta(4)$

To show that $$\sum\limits_{k=2}^{\infty} \frac{(-1)^{k}}{k^{2}} \, \left(1+\frac{1}{2}+...+\frac{1}{k}\right) \cdot \left(1+\frac{1}{2}+...+\frac{1}{k-1}\right) = \frac{3}{16}\zeta(4).$$ I came ...
24
votes
1answer
787 views

Which harmonic numbers have prime numerators?

Let $$1+\frac12+\frac13+\cdots+\frac1n=\frac{q}{p}$$ with $(p,q)=1$ and $q$ is a prime number. (I) Prove or disprove that the quantity of $n$ is limited. (II) Determine all $n$ satisfying the ...
23
votes
4answers
1k views

How find this $\sum_{n=1}^{\infty}\frac{H^3_{n}}{n+1}(-1)^{n+1}$

How Find this sum $$I=\sum_{n=1}^{\infty}\dfrac{H^3_{n}}{n+1}(-1)^{n+1}$$ where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$ My idea: since $$\dfrac{1}{n+1}(-1)^{n+1}=-\int_{-1}^{0}x^ndx$$...
23
votes
4answers
758 views

How find this sum $I_n=\sum_{k=0}^{n}\frac{H_{k+1}H_{n-k+1}}{k+2}$

$$I_n=\sum_{k=0}^{n}\dfrac{H_{k+1}H_{n-k+1}}{k+2}$$ where $$H_{n}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$$ my try:since $$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}+\dfrac{\left(1+\dfrac{1}{2}\...
23
votes
1answer
2k views

Formula for the harmonic series $H_n = \sum_{k=1}^n 1/k$ due to Gregorio Fontana.

My question was inspired by this stackexchange question. For the last 90 minutes I have been trying to prove this formula due to Gregorio Fontana: $$H_n = \gamma + \log n + {1 \over 2n} - \sum_{k=2}^\...
22
votes
2answers
2k views

A Gift Problem for the Year 2018 [duplicate]

We had this problem in exam class yesterday on Combinatoric and it was supposed to be the new year gift from our teacher. The exercise was entitled A Gift Problem for the Year 2018 Problem: ...
21
votes
1answer
1k views

$-4\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5)=S$

EDIT: Due to the solution below, I edited the answer of the post. Thanks!!!! Hi I am trying to calculate the infinite double sum $$ S:=\sum_{j,k=1}^\infty \frac{H_j(H_{k+1}-1)}{jk(k+1)(j+k)}=-4\...
21
votes
1answer
55k views

Is there a partial sum formula for the Harmonic Series? [duplicate]

There is a partial sum formula for $$\sum_{x=1}^n x^1 = \frac{n(n+1)}{2}$$ and even one when the exponent of $x$ is $0$: $$\sum_{x=1}^n x^0 = n$$ but I cannot find one for exponent $-1$: $$\sum_{x=1}^...
19
votes
5answers
9k views

Is there any formula for the series $1 + \frac12 + \frac13 + \cdots + \frac 1 n = ?$

Is there any formula for this series? $$1 + \frac12 + \frac13 + \cdots + \frac 1 n .$$
19
votes
2answers
699 views

Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$

How can I prove that $$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$ I think this post can help me, but I'm not sure.
19
votes
4answers
839 views

A closed form of $\sum_{n=1}^\infty\left[ H_n^2-\left(\ln n+\gamma+\frac1{2n} \right)^2\right]$

The series of squares of harmonic numbers $$ \sum_{n=1}^\infty H_n^2 \tag1 $$ is divergent since $\displaystyle \lim_{n \to \infty} H_n^2 \ne 0$, actually from the classic result (6.3.18), $$ H_n=\ln ...
18
votes
2answers
624 views

Closed form of $\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$

Find the closed form of $$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$$ I know the closed form for smaller powers like $2, 3$ exists, but I'm not sure if there is a closed form for this variant....
17
votes
2answers
648 views

Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$

Finding the closed form of: $$\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$$ where, $\displaystyle H_n^{(2)} = \sum\limits_{k=1}^{n}\frac{1}{k^2}$ It appears when we try to determine the ...
17
votes
1answer
342 views

A generalization of the product of harmonic numbers to non-integer arguments

This question is somewhat related to one of my previous questions: Fibonorial of a fractional or complex argument. Recall the definition of harmonic numbers: $$H_n=\sum_{k=1}^n\frac1k=1+\frac12+\,...\...
16
votes
5answers
480 views

A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$

Let $H_q$ denote harmonic numbers (generalized to a non-integer index $q$): $$H_q=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+q}\right)=\int_0^1\frac{1-x^q}{1-x}dx=\gamma+\psi(q+1),\tag1$$ where $\psi(z)=\...
15
votes
6answers
21k views

Simple proof of showing the Harmonic number $H_n = \Theta (\log n)$

Consider the partial sum of the divergent Harmonic series $H_n = \sum\limits_{k = 1}^{n}\frac{1}{k}$. I recently saw a question which required finding out the asymptotic bounds of $H_n$. Now, I could ...
15
votes
2answers
12k views

Simple proof Euler–Mascheroni $\gamma$ constant

I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n \in \mathbb{N}} = \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ converges. At first I want to know if my answer is OK. ...
15
votes
2answers
413 views

Prove that neither $A$ nor $B$ is divisible by $5$

Let the sum $$ {1+ \frac12 + \frac13 + \frac 14+ \dots +\frac1{99} + \frac 1{100}}$$ be written as $\frac AB$, where $A$ and $B$ are positive integers with no common factors. Show that neither $A$ ...
15
votes
0answers
186 views

Consecutive prime numerators of harmonic numbers?

Let $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{a}{b}$$ and let $a$ and $b$ are coprime, $h_{n}=a$. $h_{n}$ is prime for $$n=2,3,5,8,9,21,26,41,56,62,69,79,89,91,122,127,143,...
14
votes
5answers
2k views

Are we guaranteed that the harmonic series minus infinite random terms always converge?

Consider the known harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ and modify it as follows $$\sum_{n=1}^\infty a_n\frac{1}{n}$$ where $$a_n \sim \operatorname{Bern} \left({\frac{1}{2}}\right)$$ i.e. ...
14
votes
6answers
750 views

Proving $\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$

I've been trying to prove $$\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$$ I've tried perturbation and inversion but still nothing. I've even tried expanding the sum to try and find ...
14
votes
3answers
216 views

How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+…+\frac{1}{n}\right)$

How to evaluate the sum: $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$$ Can anyone help me,I really appreciate it.
14
votes
8answers
3k views

prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction

How to prove $$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$ by Mathematical induction,n$\ge $1
14
votes
3answers
670 views

Prove A New Method For Finding Primes.

I was recently messing around with harmonic numbers - that is $H_x = \sum^x_{k=1}\frac{1}{k}$ - and I was thinking what would happen if you applied Gauss' trick to the sum, which is to add up the ...
14
votes
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. ...
14
votes
2answers
556 views

Sum $\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$

I would like to seek your assistance in computing the sum $$\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$$ I am stumped by this sum. I have tried summing the residues of $\displaystyle f(z)=\frac{\pi\...
14
votes
1answer
242 views

Evaluating $1-\frac12-\frac13+\frac14+\frac15+\frac16-\cdots$

What is the value of $$S=1-\frac12-\frac13+\frac14+\frac15+\frac16-\cdots$$ where the sign alternates over the triangular numbers? To start, it is easy to prove convergence. The sum of each set of ...
14
votes
1answer
225 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 ...
13
votes
2answers
818 views

Partial sum of the harmonic series between two consecutive fibonacci numbers

I was playing around with some calculations and I noticed that the partial sum of the harmonic series: $$s_n=\sum_{k=F_n}^{F_{n+1}}\frac 1 k$$ where $F_n$ and $F_{n+1}$ are two consecutive Fibonacci ...
13
votes
3answers
1k views

Harmonic Numbers series I

Can it be shown that \begin{align} \sum_{n=1}^{\infty} \binom{2n}{n} \ \frac{H_{n+1}}{n+1} \ \left(\frac{3}{16}\right)^{n} = \frac{5}{3} + \frac{8}{3} \ \ln 2 - \frac{8}{3} \ \ln 3 \end{align} where $...
13
votes
3answers
840 views

Why is there no general form for the harmonic numbers?

The Harmonic numbers $H_n$ are given by the sum of the reciprocals of the natural numbers up to a given $n$, ie: $H_1 = 1$ $H_2 = 1 + 1/2 = 3/2$ $H_3 = 1 + 1/2 + 1/3 = 11/6$ $H_n$ for noninteger $...
13
votes
4answers
191 views

Showing $\sum_{k=1}^{nm} \frac{1}{k} \approx \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{m} \frac{1}{k}$

Since $\log(nm) = \log(n) + \log(m)$, and $\sum_{k=1}^n \frac{1}{k} \approx \log n$ for large $n$, we would expect that $$\sum_{k=1}^{nm} \frac{1}{k} \approx \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{m}...