# How to see the divergence of a series is not faster than some order? [closed]

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

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – David Roberts, user44191, Wojowu, Pietro Majer
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 $f(j) \leq f(x), j-1 \leq x \leq j$ – Conrad Mar 23 at 12:51
• 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. – KConrad Mar 23 at 14:50