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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.

6 votes
1 answer
123 views

By 'easy' I mean avoiding heavy notation. Ideally we only need numbers and multiplication signs. You can use algebra implicitly, but try to keep things elementary school level. Two examples of what I ...
Daron's user avatar
  • 11.9k
2 votes
0 answers
110 views

Let $H_n$ be the $n$th harmonic number defined by: $$ H_{n} := \sum_{k=1}^{n} \frac{1}{k} .$$ Moreover, define $S$ as the limit of the following continued fraction: $$ S := \cfrac{1}{H_{1}+ \cfrac{1}{...
Max Lonysa Muller's user avatar
2 votes
1 answer
192 views

Notation: Pochhammer symbol $(x)_{n}=\dfrac{\Gamma(x+n)}{\Gamma(x)}$, Generalized Harmonic numbers $H_n^{(r)}\displaystyle=\sum_{k=1}^{n}\dfrac{1}{k^r}$. Context: I tried to decompose $\displaystyle\...
ZCos666's user avatar
  • 31
2 votes
0 answers
76 views

Short primer on two functions which motivate my question. The generalized harmonic numbers are defined as \begin{align} H_n^{(p)}:=\sum_{k=1}^n\frac{1}{k^p}. \end{align} They admit the (ordinary) ...
Eli Bartlett's user avatar
  • 2,873
0 votes
0 answers
77 views

I am trying to approximate multi-dimensional partial harmonic series $H_n(t)$ to hopefully arrive at a generalized formula similar to the Euler-Maclaurin formula for the 1-dimensional case. Any advice ...
Glacier's user avatar
  • 157
0 votes
0 answers
102 views

Motivation: As a personal side project I have been working with an inclusion-exclusion formulation that is counting weighted power’s $x^a$ between consecutive squares $[n^2, (n+1)^2]$. The function $f(...
Glacier's user avatar
  • 157
-3 votes
1 answer
105 views

I encountered three sequences. The first sequence is explicitly defined using Harmonic numbers $H_m$: $$\text{Seq 1: } \quad \frac{1}{8}, -\frac{3}{16}, \frac{11}{48}, -\frac{25}{96}, \dots$$ The ...
Brendan's user avatar
11 votes
2 answers
719 views

I would like to solve the following differential equation $$\frac{1}{x+1} = \sum_{n=1}^{\infty} \frac{f^{(n)}(x)}{n!}.$$ Does anyone know how to solve this, or if a solution even exists? My guess is ...
Arthur Calabria's user avatar
2 votes
0 answers
83 views

When I was evaluating this monstrous integral $$ \int_0^{\frac{\pi}{2}} x^3 \ln^2 \left(\sin x\right) \, \mathrm{d}x $$ I managed to reduce it using the fact that $$ \ln^2\left(\sin x\right) = \frac{\...
imp_ractical's user avatar
4 votes
0 answers
101 views

I have been looking at sums with binomial coefficients in their denominator. These are extensions of Apery's series, which he used in his proof of the irrationality of $\zeta(3)$. This weekend I ...
aaron's user avatar
  • 739
9 votes
2 answers
278 views

how to find the following series: $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j ...
Wessel's user avatar
  • 93
0 votes
0 answers
149 views

Let $$ H_n = \sum_{k=1}^n \frac{1}{k}, \qquad n \ge 1, $$ and for a fixed parameter $r \in (0,1]$, define a sequence $(a_j)_{j\ge 1}$ by $$ a_j = 1 - \frac{r}{j}\bigl(H_{j+1}-1\bigr), \qquad j \ge 1. $...
Happy's user avatar
  • 1
2 votes
1 answer
146 views

How can you evaluate $\displaystyle\sum_{n=1}^\infty\frac{\overline{H}_n}{(2n+1)^6}=\sum_{n=1}^\infty\frac{\overline{H}_n}{n^6}-\frac{1}{64}\sum_{n=1}^\infty\frac{\overline{H}_{2n}}{n^6}$, where $\...
bruh Jardim's user avatar
5 votes
1 answer
231 views

The figure shown above shows the optimal way for stacking 30 blocks to get the maximum overhang. How does one verify/prove that this shape is indeed the best way to stack the blocks to achieve the ...
Anant S. Malviya's user avatar
1 vote
1 answer
177 views

I recently tried to approximate Harmonic numbers $H(n) $ for very large values of $n$ and that is when I accidentally came across this. My method of approach was to express $H(n)$ as a continuous ...
Quphine's user avatar
  • 154

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