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Is the given expression, monotonically increasing or decreasing with increasing x?

$\frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$

EDIT: This is the derivative of the prime counting function $\pi(x)$ w.r.t. x, ie., $\frac{d\pi(x)}{dx} \sim \frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$

Is the given expression, monotonically increasing or decreasing with increasing x?

$\frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$

EDIT: This is the derivative of the prime counting function $\pi(x)$ w.r.t. x, ie., $\frac{d\pi(x)}{dx} \sim \frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$ Also, you might check Bruce C. Berndt's "Ramanujan's NoteBooks Part IV" page 123, equation 10.16, which gives = sign instead of $\sim$

Is the given expression, monotonically increasing or decreasing with increasing x?

$\frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$

EDIT: This is the derivative of the prime counting function $\pi(x)$ w.r.t. x, ie., $\frac{d\pi(x)}{dx} = \frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$

Is the given expression, monotonically increasing or decreasing with increasing x?

$\frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$

EDIT: This is the derivative of the prime counting function $\pi(x)$ w.r.t. x, ie., $\frac{d\pi(x)}{dx} \sim \frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$

Is the given expression, monotonically increasing or decreasing with increasing x?

$\frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$

Is the given expression, monotonically increasing or decreasing with increasing x?

$\frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$

EDIT: This is the derivative of the prime counting function $\pi(x)$ w.r.t. x, ie., $\frac{d\pi(x)}{dx} = \frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$