Is there a generating function for $$\tag{1}\sum_{k\geq 1} H^{(k)}_n x^ k $$ I know that $$\tag{2}\sum_{n\geq 1} H^{(k)}_n x^n= \frac{\operatorname{Li}_k(x)}{1-x} $$ But notice in (1) the fixed $n$.

This question already has an answer here: How to show that $\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i}=0 $? 8 answers

Introduction This question appeared in the study (Sum of powers of Harmonic Numbers) of the sums of fourth powers of harmonic numbers $H_n = \sum _{k=1}^n \frac{1}{k}$ for $n=1,2,3,…$ and $H_0 = 0$. Let the sum in question be $$h_{0}(n) = \sum _{k=1}^n \frac{1}{k}H_{k-1}^2$$ Writing $H_{k-1} = H_{k} – 1/k$ we have $$h_{0}(n) = h_{1}(n) […]

Prove that : $\dfrac{n+1}{2} \leq 2\cdot\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}\cdots\sqrt[n]{2}$. I am unable to prove this even by induction and general method. Indeed, when I look at the question $2\cdot\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}\cdots\sqrt[n]{2}\leq n+1$, asked by me, I have received a hint as a comment to use binomial theorem and showed $$\left(1+\frac1n\right)^n=\sum_{k=0}^n{n\choose k}\frac1{n^k}\geq1+{n\choose 1}\frac1n=2.$$ So, expression becomes $$2 \cdot \sqrt{2} \cdot \sqrt[3]{2} […]

This question already has an answer here: Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$ [duplicate] 3 answers

Scroll down to the update to see what I am meaning. The Mathematica program below finds a solution to the equation: $$\sum _{n=1}^5 \frac{1}{n^x}+\sum _{n=1}^5 \frac{1}{n^y}=0$$ My question is if you can generalize this algorithm or iterated formula to partial sums of any length, not just $n=5$? The solution I have found is: $x=-2.30158037691463871425027298453 + […]

I apologize if this has been asked before, but I was unable to find any answers suitable for my needs so far. I have been studying the harmonic numbers over the last few days and I am unable to evaluate a particular sum involving fractional harmonic numbers. Here is a quick run-down of what I […]

Prove that for $n \in \mathbb{N}$ we have $\sum_{k=2}^n \frac{1}{k}\leq \ln(n) \leq \sum_{k=1}^{n-1} \frac{1}{k}$ by using Riemann integral?

This question already has an answer here: The limit of truncated sums of harmonic series, $\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}$ 10 answers

I’m looking for strategies for evaluating the following sums for given $z$ and $m$: $$ \mathcal{S}_m(z):=\sum_{n=1}^\infty \frac{H_n^{(m)}z^n}{n}, $$ where $H_n^{(m)}$ is the generalized harmonic number, and $|z|<1$, $m \in \mathbb{R}$. Using the generating function of the generalized harmonic numbers, an equivalent problem is to evaluate the following integral: $$ \mathcal{S}_m(z) = \int_0^z \frac{\operatorname{Li}_m(t)}{t(1-t)}\,dt, $$ where […]

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