# Bounding the finite sum of $\frac {\log n}{n}$

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.

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If this is homework, you should probably add the homework tag. People will still help you, but they will provide hints rather than full solutions. –  Andrew Oct 24 '12 at 0:20
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: en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula (along with knowing how to compute $\int\frac{1}{x}\log x\, dx$) –  Steven Stadnicki Oct 24 '12 at 0:22
oh sorry didn't know about the homework tag –  user45814 Oct 24 '12 at 1:16

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.

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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