Limit and number theory
- Tolaso J Kos
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Limit and number theory
Evaluate the limit:
$$\lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{k \leq n} \varphi(k)$$
where $\varphi$ is the Euler's function.
$$\lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{k \leq n} \varphi(k)$$
where $\varphi$ is the Euler's function.
Imagination is much more important than knowledge.
- Grigorios Kostakos
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Re: Limit and number theory
We use that \[\displaystyle\mathop{\sum}\limits_{k=1}^n\varphi(k)=\frac{3n^2}{\pi^2}+{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)\]
Because \[\displaystyle\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\,{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)=0\quad (*)\] we have that
\begin{align*}
\displaystyle\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\mathop{\sum}\limits_{k=1}^n\varphi(k)&=\frac{3}{\pi^2}+\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\,{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)\\
&=\frac{3}{\pi^2}+0\\
&=\frac{3}{\pi^2}\,.
\end{align*}
\((*)\) Left as an exercise.
Because \[\displaystyle\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\,{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)=0\quad (*)\] we have that
\begin{align*}
\displaystyle\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\mathop{\sum}\limits_{k=1}^n\varphi(k)&=\frac{3}{\pi^2}+\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\,{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)\\
&=\frac{3}{\pi^2}+0\\
&=\frac{3}{\pi^2}\,.
\end{align*}
\((*)\) Left as an exercise.
Grigorios Kostakos
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