- #1

Math100

- 779

- 220

- Homework Statement
- Prove that, for ## x\geq 2 ##, ## \sum_{n\leq x}\frac{d(n)}{n^2}=B-\frac{\log x}{x}+O(\frac{1}{x}) ##, where ## B ## is a constant that you should determine.

- Relevant Equations
- If ## x\geq 2 ## and ## \alpha>0, \alpha\neq 1 ##, then ## \sum_{n\leq x}\frac{d(n)}{n^{\alpha}}=\frac{x^{1-\alpha}\log {x}}{1-\alpha}+\zeta (\alpha)^2+O(x^{1-\alpha}) ##.

Proof:

Let ## x\geq 2 ##.

Observe that

\begin{align*}

&\sum_{n\leq x}\frac{d(n)}{n^2}=\sum_{d\leq x}\frac{1}{d^2}\sum_{q\leq \frac{x}{d}}\frac{1}{q^2}\\

&=\sum_{d\leq x}\frac{1}{d^2}(\frac{(x/d)^{1-2}}{1-2}+\zeta {2}+O(\frac{1}{(x/d)^{2}}))\\

&=\frac{x^{1-2}}{1-2}\sum_{d\leq x}\frac{1}{d}+\zeta (2)\sum_{d\leq x}\frac{1}{d^2}+O(x^{1-2})\\

&=\frac{x^{1-2}}{1-2}(\log {x}+C+O(\frac{1}{x}))+\zeta {2}(\frac{x^{1-2}}{1-2}+\zeta {2}+O(x^{-2}))+O(x^{1-2}))\\

&=\frac{x^{1-2}\log {x}}{1-2}+C\cdot \frac{x^{1-2}}{1-2}+O(x^{-2})+\zeta (2)\cdot \frac{x^{1-2}}{1-2}+\zeta (2)^2+O(x^{-2})+O(x^{1-2})\\

&=\frac{x^{1-2}\log {x}}{1-2}+\zeta (2)^2+O(x^{1-2}).\\

\end{align*}

Thus ## \sum_{n\leq x}\frac{d(n)}{n^2}=B-\frac{\log {x}}{x}+O(\frac{1}{x}) ##, where ## B=\zeta (2)^2 ##.

Therefore, ## \sum_{n\leq x}\frac{d(n)}{n^2}=\zeta (2)^2-\frac{\log {x}}{x}+O(\frac{1}{x}) ## for ## x\geq 2 ##.

Let ## x\geq 2 ##.

Observe that

\begin{align*}

&\sum_{n\leq x}\frac{d(n)}{n^2}=\sum_{d\leq x}\frac{1}{d^2}\sum_{q\leq \frac{x}{d}}\frac{1}{q^2}\\

&=\sum_{d\leq x}\frac{1}{d^2}(\frac{(x/d)^{1-2}}{1-2}+\zeta {2}+O(\frac{1}{(x/d)^{2}}))\\

&=\frac{x^{1-2}}{1-2}\sum_{d\leq x}\frac{1}{d}+\zeta (2)\sum_{d\leq x}\frac{1}{d^2}+O(x^{1-2})\\

&=\frac{x^{1-2}}{1-2}(\log {x}+C+O(\frac{1}{x}))+\zeta {2}(\frac{x^{1-2}}{1-2}+\zeta {2}+O(x^{-2}))+O(x^{1-2}))\\

&=\frac{x^{1-2}\log {x}}{1-2}+C\cdot \frac{x^{1-2}}{1-2}+O(x^{-2})+\zeta (2)\cdot \frac{x^{1-2}}{1-2}+\zeta (2)^2+O(x^{-2})+O(x^{1-2})\\

&=\frac{x^{1-2}\log {x}}{1-2}+\zeta (2)^2+O(x^{1-2}).\\

\end{align*}

Thus ## \sum_{n\leq x}\frac{d(n)}{n^2}=B-\frac{\log {x}}{x}+O(\frac{1}{x}) ##, where ## B=\zeta (2)^2 ##.

Therefore, ## \sum_{n\leq x}\frac{d(n)}{n^2}=\zeta (2)^2-\frac{\log {x}}{x}+O(\frac{1}{x}) ## for ## x\geq 2 ##.