- #1

Math100

- 756

- 205

- Homework Statement
- Use Euler's summation formula to prove that, for ## x\geq 2 ##,

## \sum_{n\leq x}\frac{ \log {n}}{n^3}=A-\frac{ \log {x}}{2x^2}-\frac{1}{4x^2}+O\frac{ \log {x}}{x^3} ##, where ## A ## is a constant.

- Relevant Equations
- None.

Proof:

Let ## x\geq 2 ##.

Then ## \frac{d}{dt}(\frac{ \log {t}}{t^3})=\frac{1-3\log {t}}{t^4} ##.

By Euler's summation formula, we have that

## \sum_{n\leq x}\frac{ \log {n}}{n^3}=\int_{1}^{x} \frac{\log {t}}{t^3}dt+\int_{1}^{x} (t-[t])(\frac{1-3\log {t}}{t^4})dt+(x-[x])\frac{log {x}}{x^3} ##

## =\frac{-\log {x}}{2x^2}-\frac{1}{4x^2}+\frac{1}{4}+(\int_{1}^{\infty}-\int_{x}^{\infty})(t-[t])\frac{1-3\log {t}}{t^4}dt+O(\frac{log {x}}{x^3}) ##.

Thus

\begin{align*}

&\left | \int_{x}^{\infty}(t-[t])\frac{1-3\log {t}}{t^4}dt \right |\leq 2\int_{x}^{\infty}\frac{\log {t}}{t^4}dt\\

&=2\cdot (\frac{1+3\log {x}}{9t^3})\\

&=O(\frac{\log {x}}{x^3}).\\

\end{align*}

Therefore, ## \sum_{n\leq x}\frac{\log {n}}{n^3}=A-\frac{\log {x}}{2x^2}-\frac{1}{4x^2}+O(\frac{\log {x}}{x^3}) ##, where ## A ## is a constant.

Let ## x\geq 2 ##.

Then ## \frac{d}{dt}(\frac{ \log {t}}{t^3})=\frac{1-3\log {t}}{t^4} ##.

By Euler's summation formula, we have that

## \sum_{n\leq x}\frac{ \log {n}}{n^3}=\int_{1}^{x} \frac{\log {t}}{t^3}dt+\int_{1}^{x} (t-[t])(\frac{1-3\log {t}}{t^4})dt+(x-[x])\frac{log {x}}{x^3} ##

## =\frac{-\log {x}}{2x^2}-\frac{1}{4x^2}+\frac{1}{4}+(\int_{1}^{\infty}-\int_{x}^{\infty})(t-[t])\frac{1-3\log {t}}{t^4}dt+O(\frac{log {x}}{x^3}) ##.

Thus

\begin{align*}

&\left | \int_{x}^{\infty}(t-[t])\frac{1-3\log {t}}{t^4}dt \right |\leq 2\int_{x}^{\infty}\frac{\log {t}}{t^4}dt\\

&=2\cdot (\frac{1+3\log {x}}{9t^3})\\

&=O(\frac{\log {x}}{x^3}).\\

\end{align*}

Therefore, ## \sum_{n\leq x}\frac{\log {n}}{n^3}=A-\frac{\log {x}}{2x^2}-\frac{1}{4x^2}+O(\frac{\log {x}}{x^3}) ##, where ## A ## is a constant.