Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I'm completely lost in my class on Additive Number Theory.

I've been trying to show that there exists a constant B such that

$$\sum_{n \le x}\frac{\log n}{n} = \frac{1}{2}\log ^2x + B + O\left(\frac{\log x}{x}\right) $$

I've honestly been working on it all night and can't come up with a good way to approach the problem!! Help!!!

The foreign O notation is really messing with me.

share|cite|improve this question
You really need to get comfortable with the $O$ notation if you're going to do number theory. – Robert Israel Oct 24 '12 at 0:20
Big hint: You shouldn't need much more than the Euler-Maclaurin formula: (along with knowing how to compute $\int\frac{1}{x}\log x\, dx$) – Steven Stadnicki Oct 24 '12 at 0:22
up vote 3 down vote accepted

Abel's partial summation technique: \begin{align*} \sum_{n=1}^{N} a(n) f(n) & = \sum_{n=1}^{N} f(n) (A(n)- A(n-1)) = \sum_{n=1}^{N} A(n) f(n) - \sum_{n=1}^{N} A(n-1) f(n)\\ & = \sum_{n=1}^{N} A(n)f(n) - \sum_{n=0}^{N-1} A(n) f(n+1)\\ & = A(N)f(N) - A(0) f(1) - \sum_{n=1}^{N-1} A(n) (f(n+1)-f(n)) \end{align*} (The above is nothing but the discrete version of integration by parts).

$$\sum_{n=1}^{N} a(n) f(n) = \int_{1^-}^{N^+} f(t) d(A(t)) = f(t) A(t) \rvert_{1^-}^{N^+} - \int_{1^-}^{N^+} A(t) f'(t) dt$$ (The second integral can be interpreted as a Riemann-Stieltjes integral.)

Consider the sum $\displaystyle \sum_{n \leq N} \frac{\log(n)}n$. Choose $a(n) = 1$ and $f(n) = \frac{\log(n)}n$ and you will get what you want.

share|cite|improve this answer
Hey Marvis, thanks for the response! Ya I missed the class earlier this week on partial summation - looks like I should go review! But thanks again, appreciated!! – user45814 Oct 24 '12 at 1:20

Your Answer


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.