How does one use Euler Maclaurin to compute asymptotics for sums like $ \sum_{\substack{n\le x \\ (n,q)=1}} \frac{1}{\sqrt{n}} \quad \text{or} \quad \sum_{\substack{n\le x \\ (n,q)=1}} \frac{\log n}{\sqrt{n}}?$
Euler Maclaurin summation examples?
2 Answers
You can get asymptotics using Abel summation (see Tenenbaum's Introduction to Analytic and Probabilistic Number Theory, section 1.0).
We can let $b(x)=\frac{1}{\sqrt{x}}$ and define $a_n$ to be $1$ if $(n,q)=1$ and 0 otherwise. Then $ \sum_{\substack{n\le x \\ (n,q)=1}} \frac{1}{\sqrt{n}} =\sum_{n \le x} a_n b(n) =A(x)b(x) - \int_1^x A(t) b'(t) \, dt = \frac{A(x)}{\sqrt{x}} + \frac{1}{2} \int_1^x A(t) t^{-3/2} \, dt $ where $ A(x) = \sum_{n \le x} a_n = \frac{x}{q} \phi(q) + O(\phi(q)).$ From this, we can determine that $\sum_{\substack{n\le x \\ (n,q)=1}} \frac{1}{\sqrt{n}} = 2 \sqrt{x} \frac{\phi(q)}{q} + O( \phi(q) ). $ I am sure better bounds on the error terms are possible, but this might be good enough for what you want to do with it.
We can actually get an additional term using the Wiener-Ikehara theorem.
Introduce the Dirichlet Series $A(s)$ whose terms are given by the indicator $(n, q) =1$ times $1/\sqrt{n}$. We have $ A(s) = \sum_{(n,q)=1} \frac{1/\sqrt{n}}{n^s} = \sum_{(n,q)=1} \frac{1}{n^{s+\frac{1}{2}}} = \prod_{p\nmid q} \frac{1}{1-\frac{1}{p^{s+\frac{1}{2}}}} = \zeta\left(s+\frac{1}{2}\right) \prod_{p\mid q} \left(1-\frac{1}{p^{s+\frac{1}{2}}} \right),$ where $p$ ranges over the primes. Furthermore, introduce $ B(s) = A(s) - \zeta\left(\frac{1}{2}\right) \prod_{p\mid q} \left(1-\frac{1}{\sqrt{p}}\right)$ This Dirichlet series differs from $A(s)$ in its constant term and converges in $\mathfrak{R}(s) \ge \frac{1}{2}.$ It has a simple pole at $\frac{1}{2}$ and is zero at $s=0.$ Wiener-Ikehara applies to $B(s)$, giving
$\sum_{k\le n,(k,n)=1} \frac{1}{\sqrt{k}} - \zeta\left(\frac{1}{2}\right) \prod_{p\mid q} \left(1-\frac{1}{\sqrt{p}}\right) \sim \prod_{p\mid q} \left(1-\frac{1}{p}\right) \frac{\sqrt{n}}{1/2} = 2 \frac{\phi(q)}{q} \sqrt{n}.$ We construct the zero at $s=0$ because we are actually working with the Mellin-Perron type integral $\int_{1-i\infty}^{1+i\infty} B(s) n^s \frac{ds}{s}$ and need to cancel the pole at $s=0$.
The conclusion is that $ \sum_{k\le n,(k,n)=1} \frac{1}{\sqrt{k}} \sim 2 \frac{\phi(q)}{q} \sqrt{n} + \zeta\left(\frac{1}{2}\right) \prod_{p\mid q} \left(1-\frac{1}{\sqrt{p}}\right)$ The numerics of this approximation are excellent even for small values of $n$.