For $d(n)$ denoting the number of divisors of an integer n, and $\sigma(n)=\sum_{d\mid n}d$ the sum of said divisors, it is known that ( eg Hardy and Wright's Intro to number theory )

$$\sum_{n\leq x}d(n)=x\log (x)+O(x) \text{, for } x\to \infty$$ and $$\sum_{n\leq x}\sigma(n)=\pi x^2/12+O(x\log (x)) \text{, for } x\to \infty$$.

Moreover, it is stated in the aforementioned book that 'assuming some theorems of Ramanujan', we may show that

$$\sum_{n\leq x}d^k(n)=x\log^{2^k-1} (x)+o(x\log^{2^k-1}) \text{, for } x\to \infty$$

After some search of these 'Theorems', I have not found something I can use to verify the above result. Nevertheless, I may now state my question in context:

How does the sum $$\sum_{n\leq x}\sigma^k(n) $$ behave for $x\to \infty$?


It can be proved (see, for instance Apostol's Introduction to Analytic Number Theory, Ch 3), that for $\alpha >0$, $\alpha \neq 1$ $$\sum_{n \leq x} \sigma^{\alpha} (n) = \frac{\zeta ( \alpha + 1)}{\alpha+1} x^{\alpha +1} + O \big( x^{\max \{ 1 , \alpha \} } \big)$$ Where $\zeta ( \cdot )$ denotes the Riemann zeta function.


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