what is asymptotic behavior of $\sum \frac {1}{\sqrt[\alpha] k}$

Question

Asymptotic behavior of $$\sum \frac {1}{\sqrt[\alpha] k}$$

for $\alpha=1$? is $\ln k$ what about $\alpha > 1$ ?

the suggested link is for $\alpha > \frac{1}{2}$ my question is about $ 0< \alpha < \frac{1}{2}$

Answer 1

Outline: try a comparison between series and integral, using the fact that for $\alpha > 0$ the function $f\colon x\in(0,\infty)\mapsto \frac{1}{x^{\frac{1}{\alpha}}}$ is decreasing.

• Specifically, for $a>0$ (and $\alpha\neq 1$), the sum will behave as $$ \int_{1}^n \frac{dx}{x^{1/\alpha}} \operatorname*{\sim}_{n\to\infty} \frac{n^{1-\frac{1}{\alpha}}}{1-\frac{1}{\alpha}} $$ as long as you can ignore the constant term in the integration and the comparison between sum and integral. It is clear that this will only happen when both the integral and the sum diverge, i.e. for $\frac{1}{\alpha} < 1$ here (or, equivalently, $\alpha > 1$).

But anyway, for $\alpha < 1$ the sum converges (it is a $p$-series), and you simply have $$ \sum_{k=1}^n \frac{1}{k^{\frac{1}{\alpha}}}\operatorname*{\sim}_{n\to\infty}\sum_{k=1}^\infty \frac{1}{k^{\frac{1}{\alpha}}}. $$

• For the case $\alpha=1$, this is the partial sum $H_n$ of the Harmonic series, whose asymptotic behavior is well-known: $$ H_n \operatorname*{\sim}_{n\to\infty} \ln n $$

Answer 2

If $\alpha=1 $ we have the well known asymptotic for the Harmonic numbers $$\sum_{k=1}^{n}\frac{1}{k}=H_{n}\sim\log\left(n\right) $$ if $\alpha>1$ we have, using Abel's summation, $$S=\sum_{k=1}^{n}\frac{1}{k^{1/\alpha}}=\sum_{k=1}^{n}1\cdot\frac{1}{k^{1/\alpha}} $$ $$=n^{1-1/\alpha}+\frac{1}{\alpha}\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{t^{1/\alpha+1}}dt $$ and since $\left\lfloor t\right\rfloor =t+O\left(1\right) $ we have $$S=n^{1-1/\alpha}+\frac{1}{\alpha}\int_{1}^{n}\frac{1}{t^{1/\alpha}}dt+O\left(\int_{1}^{n}\frac{1}{t^{1/\alpha+1}}dt\right) $$ $$=\color{red}{\frac{\alpha n^{1-1/\alpha}}{\alpha-1}+O_{\alpha}\left(1\right).}$$ If $0<\alpha<1$ the series converges to $\zeta(1/\alpha)$.