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

Prove by induction: (1n+1+1n+2+…12n)2<12,n≥1\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1

Prove by induction: (1n+1+1n+2+…12n)2<12,n≥1 For n=1 inequality holds. For n=k (1k+1+1k+2+…12k)2<12 For n=k+1 (1k+2+1k+3+…12k+2)2<12 I don’t know how to prove this. Is it better to use direct method or contradiction? Answer I have Cauchy-Schwarz inequality prove your inequality we have 1n+1+1n+2+⋯+1n+n≤√n(1(n+1)2+1(n+2)2+⋯+1(n+n)2)<√n(1n(n+1)+1(n+1)(n+2)+⋯+1(n+n−1)(n+n))=√n(1n−12n)=√22 EDIT If you must use induction, I think you must induction this following Stronger … 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

Prove $\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx$

How to prove $$\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx\tag{1}$$ Here is how I came up with this relation: In this solution @Kemono Chen elegantly proved $$\int_0^a\frac{\ln(1+ax)}{1+x^2}dx=\int_0^1\frac{a\ln(1+a^2x)}{1+a^2x^2}dx=\frac12\arctan a\ln(1+a^2)\tag{2}$$ and while trying to prove the identity in (2) starting from RHS, I ended up with the relation in (1). So any straightforward method to prove (1)? Plus any good applications of (1)? … Read more

Compute ∫10ln2(1+x)Li2(−x)xdx\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx

How to prove ∫10ln2(1+x)Li2(−x)xdx=4Li5(12)+4ln2Li4(12)−12532ζ(5)−18ζ(2)ζ(3)+74ln22ζ(3)−23ln32ζ(2)+215ln52 This integral was nicely computed by Cornel here in page 5 using tricky manipulation. Another form of the integral, after subbing and applying integration by parts is ∫10ln2(1−x)Li2(x/2)xdx My question is how to evaluate any of these integrals in a different way? Thanks Answer Finally I got the idea: Starting with … 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