Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Guo and Qi recently discovered sharp bounds for the harmonic numbers (qq.v. doi:10.1016/j.amc.2011.01.089). For example, they show that $$H_n < \ln(n) + \frac{1}{2n} + \gamma - \frac{1}{12n^2+\frac{6}{5}},$$ where $H_n$ is the $n$th harmonic number and $\gamma$ is the Euler-Mascheroni constant.

My question: are there any similar bounds on the generalized harmonic numbers? By "generalized", I mean $H_{n,r}$ where $$H_{n,r} = \sum_{k=1}^n \frac{1}{k^r}.$$

I am specifically interested in the case when $r = \frac{1}{2}$.

share|improve this question
4  
See Apostol's Introduction to Analytic Number Theory, Theorem 3.2 (b). (The proof shows that the implied constant in his Big Oh error can be taken to be 1.) Thus your sum is bounded by $2n^{1/2}+\zeta(1/2)+n^{-1/2}$ –  Stopple Feb 28 '11 at 21:03
    
Thanks Stopple, that is exactly for what I was looking. I also found a number of results by Batir and Alzer on bounding the Hurwitz Zeta function (which would of course provide a bound on the generalized harmonic numbers), however, their results are limited to $r \geq 1$. –  ESultanik Feb 28 '11 at 21:20
add comment

1 Answer

up vote 6 down vote accepted

Let $f(x)=1/\sqrt{x}$. Then by Euler-Maclaurin summation, $$ \sum_{2\leq k\leq n} \frac{1}{\sqrt{k}} = \int_1^n \frac{dx}{\sqrt{x}}+\sum_{r=0}^m\frac{(-1)^{r+1}B_{r+1}}{(r+1)!} \big(f^{(r)}(n)-f^{(r)}(1)\big)+ R$$ where $R$ is a remainder term of the form $$R=\frac{(-1)^m}{(m+1)!}\int_1^n B_{m+1}(x)f^{(m+1)}(x)dx.$$ Here $B_m$ is the $m$th Bernoulli number and $B_m(x)$ is a periodic function of period 1 that coincides with the $m$th Bernoulli polynomial on [0,1).

Taking $m=0$, for instance, since $B_1=-1/2$ and $B_1(x)=${x} - $\frac{1}{2}$ we get $$ \sum_{2\leq k\leq n} \frac{1}{\sqrt{k}} \leq 2\sqrt{n}-2 + \frac{1}{2}\Big(\frac{1}{\sqrt{n}}-1\Big)+\frac{1}{3}\int_1^n x^{-3/2} dx.$$ Now compute the integral and add back in the term $k=1$ to get an upper bound.

Sharper bounds are possible choosing $m=1,2,3,\ldots$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.