# A limit with arctan function and Euler $\varphi$ function

The following limit has stumbed upon a while. I don't have the experience to work with these kind of limit as well as these special functions like $\varphi$.

Prove that

$$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{\varphi(k) \arctan \left(\frac{k}{n} \right)}{k (n+k)} = \frac{3 \log 2}{4\pi}$$

where $\varphi$ is the Euler's totient function. How would one go to prove it?

• In what context did you encounter this limit? – Greg Martin Apr 18 '17 at 7:19
• Actually it was passed to me.. !! Question solved by the way by finding an article by Omran Kouba! No way I would have thought what he writes in the article by myself! – Tolaso Apr 18 '17 at 7:23

## 1 Answer

We can use the fact that $$\frac{\phi(k)}k = \sum_{d\mid k} \frac{\mu(d)}d,$$ where $\mu$ is the Möbius function and the sum is over all positive divisors $d$ of $k$. Then \begin{align*} \sum_{k=1}^{n} \frac{\varphi(k) \arctan \left(\frac{k}{n} \right)}{k (n+k)} &= \sum_{k=1}^{n} \frac{\arctan \left(\frac{k}{n} \right)}{n+k} \sum_{d\mid n} \frac{\mu(d)}d \\ &= \sum_{d\le n} \frac{\mu(d)}d \sum_{\substack{k\le n \\ d\mid k}} \frac{\arctan \left(\frac{k}{n} \right)}{n+k} \\ &= \sum_{d\le n} \frac{\mu(d)}{d^2} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}}, \end{align*} where we used the change of variables $k=jd$ in the last equality. For each fixed $d$, the inner sum is a Riemann sum for $\int_0^1 \frac{\arctan x}{1+x}\,dx$; since this integrand is bounded and $\sum_{d=1}^\infty \frac{\mu(d)}{d^2}$ converges absolutely to $\frac6{\pi^2}$, the dominated convergence theorem implies that \begin{align*} \lim_{n\to\infty} \sum_{d\le n} \frac{\mu(d)}{d^2} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}} &= \sum_{d=1}^\infty \frac{\mu(d)}{d^2} \lim_{n\to\infty} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}} \\ &= \sum_{d=1}^\infty \frac{\mu(d)}{d^2} \int_0^1 \frac{\arctan x}{1+x}\,dx \\ &= \frac6{\pi^2} \int_0^1 \frac{\arctan x}{1+x}\,dx, \end{align*} This reduces the problem to the evaluation $\int_0^1 \frac{\arctan x}{1+x}\,dx = \frac{\pi\log 2}8$, which is itself nontrivial but which has been answered here before (modulo an integration by parts which is natural to try).

• The integral you mention boils down to this question math.stackexchange.com/questions/172111/… . You seem to have forgotten a summation sign at the line: $$\lim_{n\to\infty} \sum_{d\le n} \frac{\mu(d)}{d^2} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}} = \sum_{d=1}^\infty \frac{\mu(d)}{d^2} \lim_{n\to\infty} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}}$$ – Tolaso Apr 18 '17 at 7:33
• @Tolaso Thank you x2! – Greg Martin Apr 18 '17 at 16:24