$$ \sum_{m=1}^{n} m^{-1+1/p} \leq Cn^{1/p} $$

For $1<p<2$, I know the LHS is divergent but I can't see its speed of divergence is not faster than $n^{1/p}$.


closed as off-topic by David Roberts, user44191, Wojowu, Pietro Majer, Joseph Van Name Mar 23 at 17:22

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  • 2
    $\begingroup$ apply the integral inequality for a decreasing function: $\sum_k^{m}f(j) \leq \int_{k-1}^{m-1} f(x)dx, k \geq 2$, $f$ continuos, decreasing, which is obtained using $f(j) \leq f(x), j-1 \leq x \leq j$ $\endgroup$ – Conrad Mar 23 at 12:51
  • $\begingroup$ Divergent series often behave like their corresponding integrals: for $0 < r < 1$, $\sum_{m=1}^n 1/n^r \sim \int_1^n (1/x^r)\,dx = n^{1-r}/(1-r) - 1/(1-r) \sim n^{1-r}$. You are asking about this with $r = 1-1/p$, where for some reason you have $p < 2$; that assumption is unnecessary. Just having $p > 1$ is adequate. $\endgroup$ – KConrad Mar 23 at 14:50