Question

If $y=\frac{logx}{x}$ , then prove by mathematical induction
$y_n=\frac{(-1{)}^n(n!)}{x^{n+1}}[logx-1-\frac{1}{2}-....-\frac{1}{n}]$

Answer 1

p (1) = $y_1=\frac{1}{x^2}$ (1 - log x)
$=\frac{(-1{)}^1,1!}{x^{1+1}}$ (log x - 1)
Assume p (n); i.e.,
$y_n=\frac{(-1{)}^n(n!)}{x^{n+1}}[logx-1-\frac{1}{2}-...-\frac{1}{n}]$
Now p (n + 1) = $y_{n+1}=\frac{d}{dx}\left(y_n\right)$
$=\frac{d}{dx}\frac{(-1{)}^2(n!)}{x^{n+1}}[logx-1-\frac{1}{2}-...-\frac{1}{n}]$ by
Differentiating as product
$y_{n+1}=(-1{)}^n(n!)$
$\times [\frac{-(n+1)}{n^{n+2}}\{logx-1-\frac{1}{2}-...\frac{1}{n}\}+\frac{1}{x^{n+1}}.\frac{1}{x}]$
$=\frac{(-1{)}^{n+1}(n+1)!}{x^{n+2}}\times [logx-1-\frac{1}{2}-...-\frac{1}{n}-\frac{1}{n+1}]$
Thus p (n + 1) also holds good.