Show that $(-1)^{n}\left( (n+1)^{\frac{1}{n+1}}-n^{\frac{1}{n}}\right)=\mathcal{O}\left(\frac{\ln(n)}{n} \right) $

Question

I would like to show:

$$(-1)^{n}\left( (n+1)^{\dfrac{1}{n+1}}-n^{\dfrac{1}{n}}\right)=\mathcal{O}\left(\dfrac{\ln(n)}{n} \right) $$

Here is my attempt

\begin{align*} (-1)^{n}\left( (n+1)^{\dfrac{1}{n+1}}-n^{\dfrac{1}{n}}\right)&=(-1)^{n}\left( \exp\left[\dfrac{\ln(n+1)}{n+1} \right]-\exp\left[\dfrac{\ln(n)}{n} \right]\right) \\ &=(-1)^{n}\left( 1+\mathcal{O}\left(\dfrac{\ln(n+1)}{n+1} \right)-1-\mathcal{O}\left(\dfrac{\ln(n)}{n} \right)\right) \\ &=(-1)^{n}\left( \mathcal{O}\left(\dfrac{\ln(n+1)}{n+1} \right)+\mathcal{O}\left(\dfrac{\ln(n)}{n} \right)\right) \\ &=(-1)^{n}\mathcal{O}\left(\dfrac{\ln(n)}{n} \right) \end{align*}

I can't prove that : $$\mathcal{O}\left(\dfrac{\ln(n+1)}{n+1} \right)+\mathcal{O}\left(\dfrac{\ln(n)}{n} \right)=\mathcal{O}\left(\dfrac{\ln(n)}{n} \right) $$

• Is my proof correct?

Answer 1

Hints

$1$. First note that

$$\dfrac{\ln(n+1)}{n+1} = O\left(\dfrac{\ln(n)}{n} \right)$$

$2$. Also you should know that

$$O(O(f(n)))=O(f(n))$$

$3$. At last

$$O(f(n))+O(f(n))=O(f(n))$$

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To prove $(1)$, you should just compute the following limit

$$\begin{align} L &= \lim_{n \to \infty} \frac{\ln [n+1]}{n+1} \frac{n}{\ln n} \\ &= \lim_{n \to \infty} \frac{n}{n+1} \lim_{n \to \infty} \frac{\ln[n+1]}{\ln n} \\ &=1 \cdot \lim_{n \to \infty} \frac{\ln[n+1]}{\ln n} \\ &= \lim_{n \to \infty} \frac{\ln[n+1]}{\ln n} \\ &= \lim_{n \to \infty} \frac{\frac{1}{n+1}}{\frac{1}{n}} \\ &= \lim_{n \to \infty} \frac{n}{n+1} \\ &= 1 \end{align}$$

where I used the L'Hoptial rule.