A series involving harmonic numbers

Does anyone know the exact value of this: ∞∑k=1(−1)kHkk or this: ∞∑k=1(−1)kH(2)kk Thanks! Thanks again for the answers! I found very interesting that the integral gives exact values up to r=3 but from 4 this integral gives not exact values: ∫0−1Li4(t)t(1−t)dt because wolfram says that “no results found in terms of standard mathematical functions” Answer … Read more

Approximating (n+1)Hn−nn2\frac{(n+1)H_n -n}{n^2}

I have the following value (n+1)Hn−nn2 But it is complicated to write, so I want to write a simple approximation to use it. I guess I can just write (n+1)Hn−nn2≈Hn−1n≈Hnn≈lnnn Is this a good approximation, or is there a clearly better one? Thanks. Answer Because γ>0.5, the approximation lognn is superior to γ+lognn despite γ=lim … Read more

Series involving harmonic numbers

Denote by Hi the i-th harmonic number. I conjecture that lim exists. I have no proof for this. I only have a vague argument. If you take H_n \approx \ln n then H_n^2 – 2\sum\limits_{i=1}^n \frac{H_i}{i} \approx \ln^2 n – 2 \int\limits_1^n \frac{\ln x}{x} dx = \ln^2 n – 2\left[\frac{\ln^2 x}{2} \right]^n_1 = -\ln^2 1 … Read more

Divergence of modified harmonic series

I am reading a paper which claims that the following series diverges: ∞∑n=21nHn−1 where Hn is the n‘th harmonic number n∑m=11m. I tried a comparison test: Since Hn<n each denominator nHn−1<n2 but that only lets me bound this from below by the convergent series ∑n1n2. How can I show that this diverges? Answer Use the … Read more

How can we one show that ∑∞i=1∑2kj=1(−1)j−12i+j=Hk?\sum_{i=1}^{\infty}\sum_{j=1}^{2k}(-1)^{j-1}{2\over i+j}=H_k?

Given the double sums (1) ∞∑i=12k∑j=1(−1)j−12i+j=Hk Where Hk is the n-th harmonic number How can one prove (1)? Rewrite (1) as ∞∑i=1(2i+1−2i+2+2i+3−⋯+2i+k) Rewrite (2) as ∞∑i=1(2(i+1)(i+2)+2(i+3)(i+4)+2(i+5)(i+6)+⋯+2(i+k)(i+k+1)) Help required, not sure what is the next step. Thank you. Answer There is some issue: if k is odd, the main term of (2) behaves like Ci, leading … Read more

Evaluating the indefinite Harmonic number integral ∫1−tn1−tdt\int \frac{1-t^n}{1-t} dt

It is well-known that we can represent a Harmonic number as the following integral: Hn=∫101−tn1−tdt The derivation of this integral doesn’t need you to derive the indefinite integral first, so now I’m wondering what the indefinite integral is and how one can derive it. According to WolframAlpha the indefinite integral is: ∫1−tn1−tdt=tn+12F1(1,n+1;n+2;t)n+1−ln(1−t)+C where 2F1(a,b;c;z) is … Read more

Find the limit lim\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n} [duplicate]

This question already has answers here: The limit of truncated sums of harmonic series, \lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}} (12 answers) Closed 6 years ago. \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n} Could you please give me a hint how to find this limit? Answer \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n} =\lim_{n\rightarrow\infty}\dfrac1n\sum_{k=1}^{n}\frac1{\dfrac kn+1} Use \lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx AttributionSource : Link , Question Author : Mirak , Answer … Read more

Binomial Harmonic Numbers

Prove this equation for 0 \leq m \leq n: \frac{1}{\binom{n}{m}}\sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} = H_n – H_{n-m} where H_k denotes the k-th harmonic number \left(~H_k := \sum_{n=1}^k \frac{1}{n}~\right). Tried to use Abels partial summation \big(\sum_{k=1}^m a_k b_k = a_m \sum_{k=1}^m – \sum_{k=1}^{m-1} (a_{k+1}-a_k)\sum_{i=1}^k b_i \big), but it leads to nowhere. Answer \newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} … Read more

Solve the recurrence relation: na_n = (n-4)a_{n-1} + 12n H_nna_n = (n-4)a_{n-1} + 12n H_n

I want to solve na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. Does anyone have an idea, what could be substituted for a_n to get an expression, which one could just sum up? We should use \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} – \frac{1}{m+1} \binom{n+1}{m+1} to simplify the result. Answer Here is a different … Read more

Is there a series approximation in terms of nn for the sum of the harmonic progression : ∑nk=011+ak\sum_{k=0}^{n}\frac{1}{1+ak}?

When a=1 the sum is given by Hn and we have : H_{n}=log(n)+γ+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4} \hspace{0.5cm}.\hspace{.1cm}.\hspace{.1cm}. Does any representation of a similar type exist for arbitrary positive a ? Answer \sum_{k=1}^{n}\frac{1}{1+ak} = \frac{1}{a}\sum_{k=1}^{n}\frac{1}{k+\frac{1}{a}}=\frac{H_n}{a}-\frac{1}{a^2}\sum_{k=1}^{n}\frac{1}{k\left(k+\frac{1}{a}\right)}=\frac{1}{a}H_n+O\left(\frac{1}{a^2}\right). AttributionSource : Link , Question Author : alex , Answer Author : Jack D’Aurizio