-2

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}$

jimjim
  • 9,675
  • 3
    http://math.stackexchange.com/questions/988710/asymptotic-behavior-of-sum-limits-k-1n-frac1k-alpha-for-alpha?rq=1. Same question, see first answer – b00n heT Jul 10 '16 at 14:18
  • 1
    I assume the bounds on your sum are $1$ and $n$, and that you are looking at the behavior when $n\to\infty$? – Clement C. Jul 10 '16 at 14:21

2 Answers2

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: $$ H_n \operatorname*{\sim}_{n\to\infty} \ln n $$
Clement C.
  • 67,323
  • Regarding the standard asymptotic behavior of $H_n$: https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence. To see a detailed example of how to perform a comparison series/integral, see "Method 2" in this answer. – Clement C. Jul 10 '16 at 14:24
  • @OP: answer edited to better address the case $\alpha < 1$. – Clement C. Jul 10 '16 at 14:40
1

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 $\zeta(1/\alpha)$.

Marco Cantarini
  • 33,062
  • 2
  • 47
  • 93
  • A quick note (I had the same thing, more or less): the asymptotics you gave are somewhat vacuous for $\alpha < 1$, since then the first term of your expansion is $o(1)$, and the partial sum is actually equivalent to its limit (as the series converges) -- which is hidden in the $O_\alpha(1)$. – Clement C. Jul 10 '16 at 14:42
  • @ClementC. you are completely right. It works only with a divergent sum, so $\alpha>1$. Thank you for pointing out. – Marco Cantarini Jul 10 '16 at 14:55