Question

Use Euler summation formula,or integration by parts in a Reimann stieltjies to show that
Summation k from one to infinity 1/k= log n- integral one to n x-[x]/x^2 dx+1

Answer 1

$$\displaystyle\sum_{k=1}^n{1\over k}=\displaystyle\int_{1}^n{1\over x}d[x]+1=$$

$$=-\displaystyle\int_{1}^n[x]dx^{-1}+n^{-1}[n]-1^{-1}[1]+1=$$

$$=\displaystyle\int_{1}^nx^{-1}dx-\displaystyle\int_{1}^nx^{-1}dx+\displaystyle\int_{1}^n{[x]\over x^2}dx+1=$$

$$=\log{n}-\displaystyle\int_{1}^n{x-[x]\over x^2}dx+1$$