OCR Output Comparison | Datalab (Chandra & Surya)

Page 1 of 1

Render HTML

236

9 Calculus and Number Theory

Proof Firstly, from (9.14) we get

$\begin{aligned} \left| P_n - \sum_{k=1}^{n} \frac{\mu(k)}{k^2} \right| &= \left| \sum_{k=1}^{n} \mu(k) \left( \frac{1}{n^2} \left\lfloor \frac{n}{k} \right\rfloor^2 - \frac{1}{k^2} \right) \right| \\ &\le \sum_{k=1}^{n} \left| \frac{1}{k^2} - \frac{1}{n^2} \left\lfloor \frac{n}{k} \right\rfloor^2 \right|. \quad (9.15) \end{aligned}$

In order to estimate the last sum above, we claim that, given natural numbers $n$ and $k$ such that $1 \le k \le n$ , we have

$\left| \frac{1}{k^2} - \frac{1}{n^2} \left\lfloor \frac{n}{k} \right\rfloor^2 \right| < \frac{2}{nk} - \frac{1}{n^2}.$

Indeed,

$\begin{aligned} \frac{n}{k} - 1 < \left\lfloor \frac{n}{k} \right\rfloor \le \frac{n}{k} & \Rightarrow \frac{n^2}{k^2} - \frac{2n}{k} + 1 < \left\lfloor \frac{n}{k} \right\rfloor^2 \le \frac{n^2}{k^2} \\ & \Rightarrow \frac{1}{k^2} - \frac{2}{kn} + \frac{1}{n^2} < \frac{1}{n^2} \left\lfloor \frac{n}{k} \right\rfloor^2 \le \frac{1}{k^2} \\ & \Rightarrow 0 \le \frac{1}{k^2} - \frac{1}{n^2} \left\lfloor \frac{n}{k} \right\rfloor^2 < \frac{2}{kn} - \frac{1}{n^2}, \end{aligned}$

as wished.

Back to (9.15), we obtain from the above estimates that

$\left| P_n - \sum_{k=1}^{n} \frac{\mu(k)}{k^2} \right| < \sum_{k=1}^{n} \left( \frac{2}{nk} - \frac{1}{n^2} \right) = \frac{2}{n} \sum_{k=1}^{n} \frac{1}{k} - \frac{1}{n}.$

Now, from L'Hôpital's rule we get

$\frac{2}{n} \sum_{k=1}^{n} \frac{1}{k} < \frac{2}{n} \left( 1 + \int_{1}^{n} \frac{1}{t} dt \right) = \frac{2}{n} (\log n + 1) \to 0$

as $n \to +\infty$ . Hence,

$\lim_{n \to +\infty} \left( \frac{2}{n} \sum_{k=1}^{n} \frac{1}{k} - \frac{1}{n} \right) = 0,$

and our previous estimates assure that