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As the title says, I'm trying to prove $$\sum_{n \le x} \frac{d(n)}{n^a}= \frac{x^{1-a} \log x}{1-a} + \zeta(a)^2+O(x^{1-a}),$$ for $x \ge 2$ and $a>0,a \ne 1$, where $d(n)$ is the number of divisors of $n$. There is a post here dealing with the case $a=1$. This is what I have done so far: \begin{align*} \sum_{n \le x} \frac{d(n)}{n^a} &= \sum_{n \le x} \frac{1}{n^a} \sum_{d \mid n} 1 = \sum_{d \le x} \sum_{\substack{n \le x \\ d \mid n}} \frac{1}{n^a} = \sum_{d \le x} \sum_{q \le x/d } \frac{1}{(qd)^a} = \sum_{d \le x} \frac{1}{d^a} \sum_{q \le x/d} \frac{1}{q^a} \\ &= \sum_{d \le x} \frac{1}{d^a} \left( \frac{(x/d)^{1-a}}{1-a} + \zeta(a) + O((x/d)^{-a}) \right) \\ &= \sum_{d \le x} \left( \frac{x^{1-a}}{d(1-a)} + \frac{\zeta(a)}{d^a} \right) + O(x^{1-a}), \end{align*} from here things start to go out of hand... I've tried using the relevant formulas from this page, but I can't get it to "fit". Any help would be appreciated.

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If you know complex analysis, try to use Perron's formula for the function $\zeta (s)^2$. –  Soarer Nov 5 '11 at 15:53
    
@Soarer: I haven't studied complex analysis... –  Carolus Nov 5 '11 at 16:09

1 Answer 1

up vote 6 down vote accepted

What you're trying to show isn't true. You should have $\zeta(a)^2$ rather than $\zeta(a)$. (See Exercise 3.3 in Apostol's number theory book in the link above.)

With that correction, you're almost there! Picking up where you left off we have, and using two of the formulas in Apostol's text in the link above,

$$ \begin{align} &\sum_{d \le x} \left( \frac{x^{1-a}}{d(1-a)} + \frac{\zeta(a)}{d^a} \right) + O\left(x^{1-a}\right) \\ &= \frac{x^{1-a}}{1-a} \sum_{d \le x} \frac{1}{d} + \zeta(a) \sum_{d \le x}\frac{1}{d^a} + O\left(x^{1-a}\right) \\ &= \frac{x^{1-a}}{1-a} \Big(\log x + O(1)\Big) + \zeta(a) \left(\frac{x^{1-a}}{1-a} + \zeta(a) + O(x^{-a})\right) + O\left(x^{1-a}\right) \\ &= \frac{x^{1-a} \log x}{1-a} + \zeta(a)^2 + O\left(x^{1-a}\right). \\ \end{align} $$

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Thanks! And, yes, that was a typo. By the way, I haven't studied any analysis and I can't say I really understand this asymptotic stuff/big Oh just yet. Could you please explain to me the following: 1) Why is $(\log x+O(1))$ equivalent to $(\log x+\gamma+O(\log x/x))$ - is it because the $O$-term now "includes" the $\gamma$? 2) How come the term $(x^{1-a} \zeta(a))/(1-a)$ disappear? As I said, I'm very much a beginner, so sorry if I'm asking really stupid questions! Thanks for the help. –  Carolus Nov 6 '11 at 5:40
    
@Carolus: 1) They aren't equivalent. An expression that is $\log x + \gamma + O(\log x/x)$ is also $\log x + O(1)$ (which is the direction I'm using), but the implication doesn't go in the other direction. Saying that an expression is $\log x + O(1)$ means that the dominant term is $\log x$ and the rest of the terms are of constant order. The expression $\log x + \gamma + O(\log x/x)$ has that property. So, in a sense, you're right to say that the $O$ term "includes" the $\gamma$. –  Mike Spivey Nov 6 '11 at 20:52
    
@Carolus: 2) It is absorbed into the $O(x^{1-a})$ expression. Since $\zeta(a)/(1-a)$ is a constant, the expression $(x^{1-a} \zeta(a))/(1-a)$ is $O(x^{1-a})$. Then the sum of two $O(x^{1-a})$ expressions is also $O(x^{1-a})$. (And no need to apologize for asking questions! Asymptotic expressions can be tricky to deal with when you first meet them.) –  Mike Spivey Nov 6 '11 at 20:57
    
Thank you very much for these explanations, I am very grateful! –  Carolus Nov 6 '11 at 22:31
    
@Carolus: You're welcome; glad it was helpful! –  Mike Spivey Nov 6 '11 at 22:31

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