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.

584 questions
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Solving limit $\lim_{n\rightarrow \infty} (\frac{1}{n+1}+ … +\frac{1}{2n})$ [duplicate]

Could anyone help me solving the following limit? $\lim_{n\rightarrow \infty} (\frac{1}{n+1}+ ... +\frac{1}{2n})$ I think it should just be 0 since we can distribute the limite inside the sum, but ...
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Prove that $\lim\limits_{n\to\infty}\frac{1}{n}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)=0$ [duplicate]

I'm preparing for exam and I have to solve this problem. Please, how do I prove that $\lim\limits_{n\to\infty}\frac{1}{n}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)=0.$ Any Theorems to guide as ...
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How to show that $\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i}=0$? [duplicate]

Show that $$\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i} =0$$ I've proved that this sequence converges (it is bounded and decreasing). NOW, I need to find a sequence that is bigger than this one and ...
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How do we rearrange the terms of the harmonic series so they add up to 0

The harmonic series converges conditionally; therefore, the series can be rearranged in any way to get different sums, but how do we rearrange in such a way that it equates to 0. Is there a trick/ ...
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Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$ by Induction

Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$ To solve it I used induction but it is leading me nowhere my attempt was as follows: Lets ...
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Elementary proof of $\sum_{k=1}^{n}\frac{1}{2k+1}$ with all odd denominators is never an integer?

For all natural number $n$, $$\frac{1}{3} + \frac{1}{5}+...+\frac{1}{2n+1}$$ is never an integer. I know that $\sum_{k=1}^{n}\frac{1}{k}$ is never an integer, but I'm not sure how to restrict that ...
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first derivative of exponential generating function of harmonic numbers

I'm trying to prove that the the first derivative of the exponential generating function of the Harmonic numbers, $H_n$, is the exponential generating function of $H_{n+1}$, but I my solution seems ...
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Justify an approximation of $-\sum_{n=2}^\infty H_n\left(\frac{1}{\zeta(n)}-1\right)$, where $H_n$ denotes the $n$th harmonic number

For integers $n\geq 1$, let $\mu(n)$ the Möbius function, see this MathWorld, and let $H_n$ the $n$th harmonic number as $H_n=1+1/2+\ldots+1/n$. Also we denote the Riemann zeta function with $\zeta(s)$...
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