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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Is there a closed-form expression for $f(x)=\sum\limits_{k=1}^\infty (\text{Ei}(i k x)+\text{Ei}(-i k x))$ assuming $x\in\mathbb{R}$?

Question: Is there a closed-form expression for $$f(x)=\sum\limits_{k=1}^\infty (\text{Ei}(i k x)+\text{Ei}(-i k x))\,,\quad x\in\mathbb{R}\tag{1}$$ where $\text{Ei}(z)$ is the exponential integral ...
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why is that $H_{k-1/2}-H_k = 2H_{2k}-H_k$?

In "Concrete Mathematics", question number 7.10 involves the factor $H_{k-1/2}-H_k$, which when I look at the answer, it states that the identities is valid: $$ \begin{align} H_{k-1/2}-H_k &...
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-2 votes
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Show that. If $H_{j}=\sum\limits_{k=1}^{j}{\frac{1}{k}}$, than $H_{2^n}\geq 1+\frac{n}{2}$ for all natural numbers.

Show that. If $H_{j}=\sum\limits_{k=1}^{j}{\frac{1}{k}}$, than $H_{2^n}\geq 1+\frac{n}{2}$ for all natural numbers.
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3 votes
3 answers
108 views

Closed form for $\sum_{n=2}^{\infty} \big{(}H_{n,n}-1\big{)} $?

It is known that $$\sum_{n=2}^{\infty}\big{(}\zeta(n)-1\big{)}=1 .$$ Many more series like these, called rational zeta series, can be evaluated in closed form. I wonder if we can also obtain similar ...
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6 votes
4 answers
361 views

Finding $\sum_{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$

I want to find the closed form of: $\displaystyle \tag*{} \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$ Where $H_{k}$ is $k^{\text{th}}$ harmonic number I tried to ...
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  • 575
4 votes
1 answer
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Are these two recursive formulas known in the literature?

First let me introduce the two recursive relations: $$\int_0^1 x^{n-1}\ln^a(1-x)dx=f(a,n),$$ where $$f(a,n)=(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}f(j,n) H_n^{(a-j)},\quad f(0,n)=\frac1n.\tag{1}$$...
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2 answers
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How this sum can be transformed into a relation with harmonic series?

I am reading and textbook and it has this line : I can not see this holds . I only get this by substituting i with (i-1): $$\sum_{i=0}^{n-1}\frac{n}{n-i} = \sum_{i=1}^n\frac{n}{n-(i-1)}$$ Can you ...
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4 votes
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Approximation of harmonic numbers and their analytical inverse.

In the same spirit as DeTemple–Wang for a series expansion of harmonic numbers, I tried to approach the problem as $$H_n\sim\frac 12 \log(n^2+n+a)+\gamma-\frac 1{b(n^2+n+a)+\Delta}\tag 1$$ hoping to ...
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18 votes
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Are these generalizations known in the literature?

By using $$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=\frac{\pi}{2^{2n+1}}\lim_{m\to \frac12}\frac{d^{2n}}{dm^{2n}}\csc(m\pi)=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\tag{a}$$ and $$\text{Li}_{a}(-z)+(-...
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2 answers
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How do I evaluate the sum $\sum_{n=1,3,5,...}^{\infty} \frac{1}{n^{4}}$? [duplicate]

I'm trying to evaluate the following sum: $\frac{480\hbar^{2}}{\pi^{4}ma^{2}} \cdot \sum_{n=1,3,5,...}^{\infty} \frac{1}{n^{4}}$ I've written out a couple partial sums for $n_{odd}$ up to $7$ : $1 + ...
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2 votes
0 answers
164 views

linear Volterra integral equation of the first kind

Solve for $f'(x)$ in $$ e^{z}-1 =-\int_{0}^{z} f'(x)\ln(1-\frac{x}{z})dx.$$ I'm fairly certain that the growth rate of $f'(x)$ is greater than polynomial. Therefore, I tried $f'(x)=e^x$ and thus $$\...
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4 votes
1 answer
82 views

Change in order of integration in an integral

I do not understand the change of order of integration ($dtdx$ to $dxdt$, RHS second line). Why the $\frac{1}{2}$ term? Why do we integrate $x$ from $t$ to $1$?
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6 votes
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Convergence of double integral involving logarithms and binomials? Can anyone present a simpler proof?

I am working through a paper I posted to arXiv and am looking to condense my results as well as seek alternative proofs for some of my intermediate results. In particular, I am looking for a shorter ...
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2 votes
1 answer
99 views

Evaluating $\sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2})$

I am seeking a closed-form (a form in terms of known special functions) to the sum $$ \sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2}). $$ Context: I am searching for closed-forms to special cases ...
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4 votes
2 answers
162 views

A sum similar to Harmonic numbers.

Consider the sum $$S_n=\sum_{r=2}^n \frac{(-1)^r\binom{n}{r}}{r-1} $$ My question is: What is the asymptotic for $S_n$ as $n\to \infty$ or find $R_n$ such that $S_n\sim R_n$ as $n\to \infty$. ...
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3 answers
79 views

Approximate $\sum\limits_{k=0}^{m-1}\frac{k}{m-k}$

How do I approximate (or better find a closed formed formula) for $\sum\limits_{k=0}^{m-1}\frac{k}{m-k}$, where $m$ is large. This looks suspiciously like a partial sum for the harmonic series, but I ...
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2 votes
1 answer
84 views

Find three integers in Harmonic Progression

It is difficult to think of many triplets which are in HP also having the condition that all of them are integers. Is there a way of systematically defining them all or a procedure to find more than a ...
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0 votes
0 answers
30 views

An identity satisfied by 'harmonic numbers' [duplicate]

Given the $harmonic\;numbers$ $$H_n=\sum_{k=1}^{n}\frac 1 k ,n\!>\!0,H_0=0$$ prove $$\sum_{k=0}^n H_k=(n+1)H_n-n$$ Also in the terminology according to the book A=B would you not agree that the $...
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9 votes
1 answer
256 views

Different ways to evaluate $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$

The following question: How to compute the harmonic series $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$$ where $H_n=\sum_{k=1}^n\frac{1}{k}$ and $H_n^{(2)}=\sum_{k=1}^n\frac{1}{k^2}$, was ...
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4 votes
2 answers
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How do I evaluate $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}H_n}{2n+1}?$

So, I am coming from this question. I managed to bring the answer to this series: $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}H_n}{2n+1}$$ where $H_n$ is the $n$th harmonic number. However, as seen in the ...
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2 votes
1 answer
57 views

Prove that this subserie of the harmonic series is convergent. [duplicate]

I tried to show the next problem but I dont see how: Let $(n_j)$ be the sequence of all naturals that do not have zero as a digit. Show that the series $ \sum_{n = 1}^{\infty} \frac{1}{n_j}$ is ...
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6 votes
4 answers
255 views

Coefficient extraction

I want to show: \begin{equation*} [z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}). \end{equation*} where $[z^n]$ means the $n$-th ...
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Is there a closed a form for $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n}\,?$

I am wondering if there is a closed form for $$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n},$$ Where $a$ is positive real. The integral representation of this sum is $$a\zeta(2)+a\int_0^1\frac{\ln(1-x)}...
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6 votes
1 answer
258 views

Trying to prove $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)$

It has been more than 7 days I have been trying to prove this following result using Harmonic Numbers Let me add this Proving $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\...
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3 votes
1 answer
92 views

The harmonic Series sequence.

Well! I was going through harmonic series from mathworld.worlfram I found harmonic numbers are really tough to calculate I was scribbling and wrote this thing $$\sum_{k = 1}^{x}H_k= \sum_{k =1}^{x}\...
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-2 votes
2 answers
64 views

nth term test for the series $\sum_{n=1}^{\infty }{1/n} $ [closed]

the n-th term test for the series $\sum_{n=1}^{\infty }{1/n}$ implies that the series converges since $\lim_{n\to\infty} 1/n =0$ but the series actually diverges. what is the error in the procedure ...
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8 votes
1 answer
304 views

Evaluating $\int_0^1\frac{\operatorname{Li}_2(x)\ln(1+x)}x\,dx$

Well, I've been trying to solve the following integral: \begin{equation*} \int_0^1\frac{\text{Li}_3(x)}{1+x}\mathrm dx, \end{equation*} where by integration by parts, making $u=\text{Li}_3(x)$ and $\...
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  • 461
1 vote
2 answers
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Hint for proving $H_n=\sum_{i=1}^n\binom{n}{i}\frac{(-1)^{i+1}}i$, where $H_i=\sum_{k=1}^i\frac1k$

I have been trying to prove the following identity: $$H_n = \sum_{i=1}^{n} \binom{n}{i}\frac{(-1)^{i+1}}{i}$$ where $$ H_i = \sum_{k=1}^{i} \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \...
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1 vote
0 answers
36 views

Asymptotic behaviour of the $\sum_{k=3}^n \frac{n!}{(n-k)! n^k k}$ [duplicate]

We can show that $\sum_{k=3}^n \frac{n!}{(n-k)! n^k k} = \Theta(\ln(n))$ using the fact that $\sum_{k=3}^n \frac{1}{k} = \Theta(\ln(n))$ and $\frac{n!}{(n-k)! n^k} = \Theta(1)$ when $k=[\sqrt{n}]$. ...
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0 votes
0 answers
218 views

Cornel's triple binoharmonic series

The following sum of series was recently proposed by Cornel, $$\sum_{n=1}^{\infty}\frac{2^{2n+1}H_n}{\displaystyle n(2n+1)^2\binom{2n}{n}}+\sum_{n=1}^{\infty}\frac{H_{n-1}^2}{\displaystyle(2n-1)^2 2^{...
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5 votes
2 answers
170 views

Integer Values of $\sum_{k=1}^n k^r . \sum_{q=1}^n \frac{1}{q^r}$

For harmonic numbers $H_n = \sum_{k=1}^n \frac{1}{k}$ we know that this sum is never integer for any $n$. The same is true for generalized Harmonic numbers: the sum $\sum_{k=1}^n\frac{1}{k^r}$ is ...
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4 votes
1 answer
264 views

Prove $\displaystyle\sum_{-\infty}^0\text H(x)=\frac12\sum_0^\infty\frac{\text H_\frac x2-\text H_\frac{x-1}2}{x!}=\frac{\text{Ei}(2)-\text{Ei}(1)}e$

The Hadamard Gamma function generalizes the factorial function to have no poles as a result of the reciprocal gamma function in its definition. It also is monotonic on $[-\infty,0]$ and $[2,\infty]$ ...
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2 votes
1 answer
198 views

changing the signs of the harmonic series so that it is still divergent

While studying Knopp's book and solutions to an interesting problem here: changing the signs of the harmonic series so that it converges? and here: Change of signs in harmonic series, the following ...
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2 votes
1 answer
86 views

Writing $ \frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}} =\frac{m}{n}\left( H_n-H_{m-n} \right) $

Given the following series, where $m,n$ are integers such that $n<m$: \begin{align} \frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}}\\ =\frac{m}{n}\left( H_n-H_{m-n} \right) \end{align} We know that ...
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  • 1,071
3 votes
4 answers
108 views

Evaluate the following : $\lim_{n \to \infty} \left\{ \frac{n!}{\prod_{k=1}^{n} H_k} \right\}^{\frac{1}{n}}$ where $H_k = \sum_{i=1}^{k} \frac{1}{i}$.

Evaluate the following limit : $$\lim_{n \to \infty} \left\{ \frac{n!}{\prod_{k=1}^{n} H_k} \right\}^{\frac{1}{n}}$$ where $$H_k = \sum_{i=1}^{k} \frac{1}{i}.$$ This question was from this; I tried ...
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  • 995
2 votes
1 answer
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Card Trick of the 'Prime Obsession' book

I saw this problem at the first chapter of the book called Prime Obsession by John Derbyshire: Take an ordinary deck of 52 cards, lying on a table, all four sides of the deck squared away. Now, with ...
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1 vote
1 answer
74 views

An identity involving the central binomial coefficient and the harmonic numbers

While doing some computational experiments, I found the following (conjectured) identity: $$2^{2n-1}\sum_{k=0}^{\infty}(-1)^{k}\frac{a_{n,k}}{2^k}=\binom{2n}{n}$$ where $a_{k}$ is defined as $$a_{n,k+...
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1 vote
0 answers
32 views

Harmonic number denominator divisibilty proof

I'm looking for the complete proof (in a textbook preferably) of harmonic number $H_n$ denominator being divisible by all primes $p\in\left[\frac{n+1}{2},n\right]$. I found one by induction in ...
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16 votes
0 answers
284 views

Show that there are no real solutions of $1 + \sum_{n = 1}^{\infty} \frac{x^n}{\prod_{k=1}^{n} H_k} = 0$ where $H_k = \sum_{i=1}^{k} \frac{1}{i}.$

[Edited] Show that there are no real solutions of $$1 + \sum_{n = 1}^{\infty} \frac{x^n}{\prod_{k=1}^{n} H_k} = 0$$ where $$H_k = \sum_{i=1}^{k} \frac{1}{i}.$$ I managed to prove this, and I want to ...
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  • 995
2 votes
0 answers
136 views

Does $\sum_{n=0}^\infty \frac{1}{nH_n^2}$ (& related) have any closed form representaions?

We know that: $$\sum \frac{1}{n^{1+\epsilon}} \quad converges \ \forall \epsilon>0 \quad \& \quad \sum \frac{1}{n} \sim \ ln(n) \tag{1}\label{asymp1}$$ Using this we define: $$\gamma_1:= \lim_{...
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1 vote
0 answers
74 views

Does $ \sum_{k=2}^{\infty}\frac{H_{k-1}}{k(k+n)}=\frac{1}{n}\sum_{k=1}^{n} \frac{H_k}{k} $?

I have been trying to do Problem 1.3.2 from the fantastic Almost Impossible Integrals book using different solution techniques. I tried using a Cauchy product expansion and I managed to simplify it ...
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1 vote
1 answer
87 views

Prove the series $\sum_{n=1}^{\infty} \frac{n^2{a_n}}{(a_1 + a_2 + ... + a_n)^{2}}$ converges [duplicate]

Prove the series $\sum_{n=1}^{\infty} \frac{n^2{a_n}}{(a_1 + a_2 + \dots + a_n)^{2}}$ converges given $\sum_{n=1}^{\infty} \frac{1}{a_n}$ convereges, $\forall a_n > 0$ This is a question that a ...
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2 votes
0 answers
94 views

Computation of $\sum_{n\geq1}(-1)^{n+1}\frac{\text{H}_{n}\zeta(1+n)}{1+n}$ in terms of zeta function

I would like to get an expression which represents the series $$\sum_{n\geq1}(-1)^{n+1}\frac{\text{H}_{n}\zeta(1+n)}{n+1}$$ in terms of $\zeta (2)$ and a series expansion containing $\zeta (n)$. Here, ...
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  • 5,571
3 votes
1 answer
110 views

How i can to calculate the harmonic partial series: $\sum_{k=1}^n\frac{\left(H_k\right)^2-H^{(2)}_k}{k}$?

I have trying to calculate $$S_n=\sum_{k=1}^n\frac{\left(H_k\right)^2-H^{(2)}_k}{k} $$ I'm trying to apply Abel's Summation, i.e.: $$ \sum_{k=m}^n a_kb_k=A_nb_n-A_{m-1}b_m-\sum_{k=m}^{n-1}A_k\left(b_{...
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  • 461
1 vote
1 answer
98 views

Upper limit for $(1+1/2)(1+1/3)\dots(1+1/n)$

I am trying to find an upper limit for the product $(1+1/2)(1+1/3)\dots(1+1/n)$. I tried using AM-GM inequality as follows: \begin{align} (1+1/2)(1+1/3)\dots(1+1/n) \leq \left(\frac{1}{n-1}\left( -1 + ...
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  • 1,047
3 votes
1 answer
184 views

Sums of harmonic series

If $$f(n, a, b)=\sum_{k_1=a+1}^{b}\sum_{k_2=a+1}^{k_1}\cdots\sum_{k_{2n}=a+1}^{k_{2n-1}}\frac{1}{k_1k_2\ldots k_{2n}}$$ $f(0, a, b)=1$ and $$g(a, b)=\sum_{n=0}^{\infty}f(n, a, b)$$ is there a way to ...
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  • 472
0 votes
5 answers
150 views

Approximating the Digamma function for small arguments

There are several ways to approximate the Digamma function $\psi(x)$ that become exact for $x\to\infty$. The simplest approximation is $$\lim_{x\to\infty}(\psi(x)-\mathrm{ln}(x))=0$$ There are other ...
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9 votes
3 answers
423 views

The Double Basel Problem [duplicate]

I have been playing with the series which I had been calling the 'Double Basel problem' for the past couple of hours $$ \sum_{n=1}^{\infty} \sum_{m=1}^\infty \frac{1}{{n^2 +m^2}}. $$ After wrestling ...
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  • 524
3 votes
0 answers
295 views

Fascinating equality

The following problem is proposed by a friend:$$4\int_0^{\pi/4}\left(\int_x^{\pi/4} (x-y)\ln(\tan (x)) \ln\left(\tan \left(y+\frac{\pi }{4}\right)\right) \textrm{d}y\right)\textrm{d}x$$ $$-\sum _{n=1}^...
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  • 21.6k
1 vote
1 answer
167 views

Why is $\lim\limits_{x\to\infty}\frac{\sum_{i=1}^x(\sum_{j=1}^i\frac1j-\ln i-\gamma)}{\sum_{i=1}^x\frac1i}=\frac12$?

$$ \mbox{Why is}\quad\lim_{x\to\infty} \frac{\sum_{i = 1}^{x}\left[\sum_{j = 1}^{i}1/j -\ln\left(i\right)-\gamma\right]}{\sum_{i = 1}^{x}1/i} = \frac{1}{2}\ ?.$$ I learnt Euler's Constant $\gamma$ ...
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