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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3
votes
2answers
74 views

Can $\int_0^\infty f (x) \, dx$ exist if $\lim_{x \to \infty} f(x)$ does not exist?

Is is possible to have a function for which $\lim_{x \to \infty} f(x)$ does not exist, but $\int_0^\infty f(x) \, dx$ exists and is finite? I think I've found an example actually, but I'm not sure it ...
7
votes
1answer
113 views

$\zeta(4)$ in terms of a series of $\zeta(3)$ and harmonic numbers

The other day I believe I found a proof that: $$\sum_{k=1}^\infty \frac{\zeta(2)-H_k^{(2)}}{k} = \zeta(3)$$ I was wondering if a general recursion like this was well known, but I couldn't find ...
2
votes
3answers
49 views

How was the harmonic number's relation to the hurwitz zeta function derived?

How was the generalized harmonics number's relation to the hurwitz zeta function derived? $H_{n,\ m} =\zeta ( m,\ 1) -\zeta ( m,\ n+1),\ \Re(m)>1$ I tried looking at the series representations for ...
1
vote
1answer
68 views

How to Evaluate the Sum $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\left(H_{n}\right)^2}{2n+1}$

How can I evaluate the Sum : $$S=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\left(H_{n}\right)^2}{2n+1}$$ where $\left(H_{n}\right)^2$ denotes a Harmonic Number Squared. The sum converges and is approximated ...
1
vote
3answers
78 views

Upper bound of $\sum_{k=1}^n \frac{1}{\sqrt{k}}$?

I am looking for an upper bound of $\sum_{k=1}^n \frac{1}{\sqrt{k}}$. Alternatively, is the sequence $\frac{1}{n\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt{k}}$ bounded? I am trying to use a Strong law of ...
5
votes
1answer
80 views

Fourier Legendre expansion of $_3F_2\left( 1,1,1;\frac32,\frac32;x\right)$

Problem: Can we obtain a closed-form Fourier-Legendre expansion of the following hypergeometric series? $$_3F_2\left( 1,1,1;\frac32,\frac32;x\right)$$ Or equivalently, how to evaluate $$I(n):=\int_0^1\...
6
votes
1answer
89 views

If $ 1+ \frac{1}{2}+\frac{1}{3}+…+\frac{1}{100}=\frac{A}{B}$ where $A$ and $B$ are coprime positive integers, then $5\nmid A$ and $5\nmid B$.

Let the sum $$1+ \frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}=\frac{A}{B}$$ where $A,B\in \mathbb{N}$ and $\gcd(A,B)=1$. Show that neither $A $ nor $B $ is divisible by $5$. My attempt: $$\begin{...
4
votes
3answers
103 views

Evaluation of double summation

I am trying to evaluate the following summation : $\sum_{m=2}^{\infty}\sum_{n=1}^{m-1}\frac{1}{(2m-1)(2n-1)(2m-2n)}$ The above summation is a partial sum of another double summation, $\sum_{m=1}^{\...
2
votes
1answer
50 views

Can these two conjectured relationships between two prime counting functions and the harmonic number function $H(x)$ be proven?

This question assumes the following definitions where $p$ is a prime and $n$ and $k$ are positive integers. (1) $\quad\pi(x)=\sum\limits_{p\le x} 1\quad\text{(fundamental prime counting function)}$ (...
4
votes
1answer
120 views

F-L expansion, $\sum _{n=0}^{\infty } \left(\frac{\binom{2 n}{n}}{4^n}\right)^2\frac{1}{(2 n+1)^4}$, $_pF_q$ and MZV

Background: I'm searching for hypergeometric representations for high level MZVs. I tried several methods to generalize known identities to higher weights and they all succeeded, except the following ...
3
votes
0answers
54 views

How to show $\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^3}$ using series manipulation?

Using integration, I managed to show that $$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^3}\tag1$$ But I would like to prove the equality using ...
0
votes
2answers
36 views

Explanation of part of a particular proof by induction that the harmonic series diverges

One proof by induction that a harmonic series diverges begins $\sum_{j=1}^{2^{n+1}}\frac{1}{j}>\frac{(n+1)+1}{2}=\frac{n+2}{2}$ so: $$\begin{aligned}\sum_{j=1}^{2^{n+1}}\frac{1}{j}&=\sum_{j=1}^{...
2
votes
4answers
73 views

Evaluating the indefinite Harmonic number integral $\int \frac{1-t^n}{1-t} dt$

It is well-known that we can represent a Harmonic number as the following integral: $$H_n = \int_0^1 \frac{1-t^n}{1-t} dt$$ The derivation of this integral doesn't need you to derive the indefinite ...
0
votes
1answer
51 views

Does the green area converge to a known constant when $n\to \infty$?

Let $n$ denote the number of the rectangles in the figure above. We know that the gray area converges to Euler-Mascheroni constant $(\gamma)$ when $n\to \infty$. I have three questions about the ...
1
vote
1answer
75 views

How to compute $\sum_{n=1}^\infty \frac{H_{2n}^2}{n^2}$?

where $H_n$ denotes the harmonic number. I can't see $$\sum_{n\geq 1} \frac{1}{n^2}\left(\int_0^1 \frac{1-x^{2n}}{1-x}\ \mathrm{d}x\right)^2$$ be of any assistance; even $$-\sum_{n\geq 1}H_{2n}^2\...
2
votes
1answer
65 views

Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$?

Here, $H_n$ denotes the harmonic number. More colloquially, is there any way to represent $$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x$$ in a nice way? The latter is corollary to the original ...
2
votes
1answer
82 views

Evaluate $\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}$.

I am looking for a closed for $$\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}.$$ I believe there is a closed form for the sum as we have seen in [1] which poses as, presumably, a more difficult sum of ...
1
vote
2answers
73 views

How to evaluate $ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}$

I am having a difficult time evaulating $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}.$$ I have tried the following relation: $$\frac{1}{2}\int_0^1 \frac{\mathrm{Li}_2(x)}{x(1-x)}\log^2{x}\ \mathrm{d}x.$$ ...
3
votes
2answers
92 views

Closed form for $\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x$

I am looking for a closed form for: $$\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x.$$ I am assuming integration by parts multiple times but I can't get anywhere with it. Any help/...
1
vote
1answer
29 views

Solution with harmonic numbers

Let $H_k = \sum_{i=1}^k \frac{1}{i} $ be the $k$-th Harmonic number. We have that $$\sum_{k=1}^n H_k = (n+1) H_{n+1}-(n+1)$$ My question is: If $$\sum_{i=1}^{p-1} C_i^s = (s+p) C_p^s -(s+p+1)$$ ...
4
votes
2answers
156 views

Quadratic Euler sums $\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \widetilde H(n)^3}{2 n+1}$

I am investigating a class of quadratic sums of weight 4 and there're only 3 sums remain. Denote $H(n)=\sum_1^n \frac{1}{k}, \widetilde H(n)=\sum_1^n \frac{(-1)^{k-1}}{k}$ the non-alternating/ ...
1
vote
1answer
118 views

Looking for closed form for $\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \mathrm{d}x$

I was evaluating and integral involving iterated logarithms when the following integral appeared: $$\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \...
4
votes
0answers
96 views

Is the generalization $\sum_{n=1}^\infty\frac{H_{\frac np}}{n^q}$ known in the literature?

I managed to derive the following generalization $$\sum_{n=1}^\infty\frac{H_{\frac np}}{n^q}=(-1)^qp \sum_{n=1}^\infty\frac{H_{pn}}{(pn)^q}-\sum_{j=1}^{q-2}(-p)^{-j}\zeta(q-j)\zeta(j+1)\tag1$$ and ...
3
votes
0answers
59 views

Harmonic number evaluated at $m + \sqrt{n}$

The function for harmonic numbers is given by $$H_x = \int_0^1 \frac{1-t^x}{1-t} dt \tag 1$$ For rational $x$, the formula is given by $$H_{x} = \frac{1}{x} + 2\sum_{k=1}^{\left\lfloor \frac{q-1}{2} \...
0
votes
1answer
40 views

When can you upper bound a summation by the Harmonic number?

I am currently studying a distributed systems lecture. I came across a relation which I was somewhat confused by. It states that the following equation $$ 1+ \sum_{r=2}^n \frac{(n-1)(1)}{(n)(r-1)} $$...
5
votes
2answers
66 views

Is $H_m - H_n$ a surjection onto $\mathbb{Q}^+$?

I was wondering whether, for each rational $q$, we may always write $$q = \sum_{k=a}^b \frac 1k$$ For some positive integers $a \leq b$. I get the feeling that this is not true (although an ...
3
votes
2answers
79 views

How to prove this identity on exponential generating function of harmonic numbers

I came across the following problem, let $N![z^N]A(z)$ denote the coefficient of an exponential generating function (EGF) $A(z)$. The EGF is similar to an ordinary generating function (OGF) $A'(z)$ ...
2
votes
2answers
63 views

How can we compute $\sum_{m=1}^\infty \frac{H_{\left(m-3/2\right)}}{m^2}$?

In the evaluation of $$\sum_{k=1}^\infty \sum_{\ell=1}^{k-1}\sum_{m=1}^{\ell-1}\frac{\delta_{k, 2\ell-2m}}{m\left(\ell-m\right)\left(k-\ell\right)}.$$ Here $\delta_{k, 2\ell-2m}$ denotes the "...
0
votes
1answer
16 views

summation of random variable into harmonic number

In our textbook "Algorithm Design" we are given an example of a deck of $n$ cards and we have to guess the correct card, and every time a card is drawn, we remember that cards so we are uniformally ...
2
votes
1answer
115 views

Quadratic Euler Sum $\sum_{n=1}^\infty \frac{(\pm1)^{n-1}}{n^2}\left(\sum_{m=1}^n \frac{(-1)^{m-1}}{2m-1}\right)^2$

Question: denote $$J_n=\sum_{m=1}^n \frac{(-1)^{m-1}}{2m-1}$$ quadratic alternating harmonic number, then how can we calculate $$\sum_{n=1}^\infty \frac{(\pm1)^{n-1}}{n^2}J_n^2\ \ \text{and}\ \ \sum_{...
1
vote
2answers
78 views

Closed form or simplification of $\sum_{k=1}^{n} \frac{x^{n-k}(1-x^k)}{k}$

Does there exist a simpler form of $$\sum_{k=1}^{n} \dfrac{x^{n-k}(1-x^k)}{k}$$ For $x \in ]0,1[$. Perhaps some combinatorial interpretation?
0
votes
3answers
109 views

Mustering $\sum_{k=1}^{\infty} \frac{(-1)^kH_{2k}}{k^2}$ with complex series

I have a series $$\sum_{k=1}^{\infty} \frac{(-1)^kH_{2k}}{k^2}$$ And I attempted to use the generating function $$\sum_{k=1}^{\infty} \frac{x^kH_k}{k^2} = \mathrm{Li}_2(1-x)\ln(1-x) + \frac{1}{2}\...
2
votes
2answers
70 views

Another series involving $\log (3)$

I will show that $$\sum_{n = 0}^\infty \left (\frac{1}{6n + 1} + \frac{1}{6n + 3} + \frac{1}{6n + 5} - \frac{1}{2n + 1} \right ) = \frac{1}{2} \log (3).$$ My question is can this result be shown ...
1
vote
1answer
95 views

Prove $\sum_{k=1}^{\infty} \frac{H_{k}^{(2)}}{2^kk} = \frac{5\zeta(3)}{8}$

$$\sum_{k=1}^{\infty} \frac{H_{k}^{(2)}}{2^kk} = \frac{5\zeta(3)}{8}$$ Can someone show this without making a generating function for $$\sum_{k=1}^{\infty} \frac{x^kH_{k}^{(2)}}{k}$$ Since that is ...
2
votes
2answers
143 views

Compute $\int_0^1 \frac{\text{Li}_2(-x^2)\log (x^2+1)}{x^2+1} \, dx$

How can we evaluate: $$\int_0^1 \frac{\text{Li}_2\left(-x^2\right) \log \left(x^2+1\right)}{x^2+1} \, dx$$ As one can see, much of this kind of polylog integrals and harmonic sums are around this site....
3
votes
2answers
100 views

Sum $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}H_{2k}}{k}$

I am particularly interested in solving $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}H_{2k}}{k}$$ Where $$H_n = \sum_{k=1}^n \frac{1}{k}$$ I can’t seem to crack it. As much as I’d love to use the ...
1
vote
1answer
31 views

Understanding this property of the digamma function: $\pi(x + \frac{1}{2}) = -\gamma - 2\ln 2 + \sum\limits_{k=1}^n\frac{2}{2k-1}$

I am reading through the Wikipedia article on the digamma function where the digamma function is compared to the Harmonic Numbers defined as $H_n = \sum\limits_{k=1}^{n} \dfrac{1}{k}$. The first ...
5
votes
2answers
68 views

Sum of series with conditional convergence

Sorry for this question, but for some reason I'm stuck on this for few hours already. Before I solved more complex ( I think ) problems, but can't solve this. The only thing I know that this series ...
0
votes
1answer
50 views

Sum of series scaled by the harmonic series

Let $\{ a_k \}$ so that $\sum_{k=1}^n a_k = A_n$. We can also assume that $A_n \rightarrow A$. It is known that $\sum_{k=1}^n \frac1k \approx \ln(n)$. Is there a way to approximate $\sum_{k=1}^n \...
0
votes
3answers
35 views

Is there a simpler expression for the sum $\sum_{i=1}^{\infty} \frac{H_{i}}{i+1}$?

Is there any way to simplify the following series? $$\frac{1}{2}(1) + \frac{1}{3}(1 + \frac{1}{2}) + \frac{1}{4}(1+\frac{1}{2}+\frac{1}{3}) + \frac{1}{5}(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) +...
1
vote
1answer
70 views

How close to an integer can the harmonic series of prime numbers be?

The harmonic series of prime numbers can be written as $\sum_{i=1}^{n}\frac{1}{p_{i}} = \frac{p_{2}p_{3} \cdots p_{n} + p_{1}p_{3} \cdots p_{n} + \ldots + p_{1}p_{2} \cdots p_{n-1}}{p_{1}p_{2} \cdots ...
2
votes
2answers
71 views

Prove $\sum_{i=1}^{n-1} \left[\frac{n}{i(i+1)} + \frac{n(n-1)}{i(i+1)} (n(H_{n-2} - H_{n-i-1}) - (i-1))) \right] =(n-1)^2$?

Apparently the following expression $$ \sum_{i=1}^{n-1} \Bigg[\frac{n}{i(i+1)} + \frac{n(n-1)}{i(i+1)} (n(H_{n-2} - H_{n-i-1}) - (i-1))) \Bigg] \\ $$ simplifies to $(n-1)^2$, where $H_i$ is ...
2
votes
0answers
117 views

Binomial Euler sums: Evaluate $\sum_{n=1}^\infty\frac1{4^n}\binom{2n}n\frac{H_n^{(s_1)}\cdots H_n^{(s_k)}}{n^s}$

Denote $$f(s;s_1,s_2,\ldots,s_k)=\sum_{n=1}^\infty\frac1{4^n}\binom{2n}n\frac{H_n^{(s_1)}\cdots H_n^{(s_k)}}{n^s}$$ Can $f(\cdots)$ always be represented as $\mathbb Q$-linear combination of ...
2
votes
5answers
119 views

Convergence of $\sum_{n=1}^{\infty}\frac{(-1)^{T_n+1}}{n},$ where $T_n$ is the $n$th Triangular number

Consider the series $$\sum_{n=1}^{\infty}\frac{(-1)^{\frac{n(n+1)}{2}+1}}{n}=1+\dfrac12-\dfrac13-\dfrac14+\dfrac15+\dfrac16-\cdots.$$ This is clearly not absolutely convergent. On the other hand, ...
5
votes
1answer
130 views

How to determine bounds on one variable in a system of inequalities?

I came across this problem whilst exploring the asymptotic behaviour (or not) of different generalised harmonic numbers. I am interested in the point of 'cross-over' between a generalised harmonic ...
0
votes
0answers
34 views

Can I get a bound on $\sum_i x_i$ given a bound on $\sum_i x_i/(1+x_i)$ and some conditions on $x_i$?

This is a follow-up question on a question I asked previously. Suppose that $x_i \geq 0$ and for all $n \geq n_0$, we have $$\sum_{i=1}^n \frac{x_i}{1+x_i} \leq \text{poly-log}(n),$$ where $\text{...
3
votes
0answers
48 views

Prove that $\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n}\zeta(n) = \gamma$

How do you prove that $$\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n}\zeta(n) = \gamma$$ where $\gamma$ is the Euler-Macheroni constant? This series kind of appeared in one of the questions I asked earlier; ...
1
vote
2answers
70 views

Can I get a bound on $\sum_i x_i$ given a bound on $\sum_i x_i/(1+x_i)$?

Suppose that $x_i \geq 0$ and for any $n \geq n_0$, we have $$\sum_{i=1}^n \frac{x_i}{1+x_i} \leq a \log n + b,$$ for some positive constants $a$ and $b$. Given this information, is it possible to ...
0
votes
1answer
24 views

Can $0$ be an element of a harmonic series?

The definition of a harmonic series is that for any $H_{k-1},H_k,H_{k+1} \in \{H_1,H_2, ... H_n\}$, $$\frac{2}{H_k} = \frac{1}{H_{k-1}} + \frac{1}{H_{k+1}}$$ If any $H_k$ is zero, then $\frac 1 {H_k}$...
6
votes
1answer
86 views

Questions regarding $\ln(x) = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}(\zeta(n,x)-\zeta(n))$. Have I found something “new”?

Introduction TL;DR I was messing around with the Taylor series for $\ln(x)$ when I ended up with the formula \begin{align} \ln(x) &= \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}(\zeta(n,x)-\zeta(n)) \\\...

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