1

I have a proof where the divergence harmonic series is shown via the Cauchy criteria.

$\epsilon := 1/2$ and $m := 2n$.

$$ a_m - a_n = (1 + \frac{1}{2} + \ldots + \frac{1}{m}) - (1+\frac{1}{2} + \ldots + \frac{1}{n}) \\ = (1 + \frac{1}{2} + \ldots + \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{2n}) - (1+\frac{1}{2} + \ldots + \frac{1}{n}) \\ = (\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n}) \geq \frac{1}{2n} + \frac{1}{2n} + \ldots + \frac{1}{2n} = \frac{n}{2n} = \frac{1}{2} = \epsilon) $$

I'm trying to understand where the $\geq$ and the $\frac{1}{2n} + \frac{1}{2n} + \ldots + \frac{1}{2n}$ comes from in the last line. Also, why should $\frac{1}{n+1}$ be larger than $\frac{1}{2n}$ if in the second line the series is progressing from $\frac{1}{n+1}$ to $\frac{1}{2n}$. In my understanding, the value should then be smaller.

Thanks for your help!

Hellstorm
  • 133
  • 2
  • 1
    Well, we have $2n\geq n+1$ so we have the reverse inequality for the reziprocals. What did you mean when referring to the second line? – Jonas Lenz Jun 06 '18 at 07:12

1 Answers1

1

Observe that if $1 \le k \le n \implies n+k \le 2n \implies \dfrac{1}{n+k} \ge \dfrac{1}{2n}$. Let $k$ varies from $1$ to $n$ and add them up. Also the number of inequalities is $2n-(n+1)+1 = n$ which explains the numerator $n$ of the fraction $\dfrac{n}{2n}$ that is $\dfrac{1}{2}$ after simplifying .

DeepSea
  • 77,651