A limit with Euler's totient function
- Tolaso J Kos
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A limit with Euler's totient function
Here is something I created.
Let $\varphi$ denote Euler’s totient function. Evaluate the limit
$$\ell = \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{k=1}^{n} \sin \left (\frac{\pi k}{n} \right) \varphi(k)$$
Let $\varphi$ denote Euler’s totient function. Evaluate the limit
$$\ell = \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{k=1}^{n} \sin \left (\frac{\pi k}{n} \right) \varphi(k)$$
Imagination is much more important than knowledge.
Re: A limit with Euler's totient function
We are quoting a theorem by Omran Kouba:
Hence the limit is $\frac{6}{\pi^3}$.
Theorem:Proof: The theorem can be found at the attachment following:
Let $\alpha$ be a positive real number and let $\{a_n\}_{n \in \mathbb{N}}$ be a sequence of positive real numbers such that
$$\lim_{n \rightarrow +\infty} \frac{1}{n^\alpha} \sum_{k=1}^{n} a_k = \ell$$
For every continuous function $f$ on the interval $[0, 1]$ it holds that
$$\lim_{n \rightarrow +\infty} \frac{1}{n^\alpha} \sum_{k=1}^{n} f \left ( \frac{k}{n} \right ) a_k = \ell \int_{0}^{1} \alpha x^{\alpha-1} f(x) \, {\rm d}x$$
Hence the limit is $\frac{6}{\pi^3}$.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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