Why does $\frac{1}{n+\ln n} ∼ \frac{1}{n}$?

Question

I have to find if the sequence $\frac{1}{n+\ln n}$ is convergent or divergent? And in the correction they've just wrote : $\frac{1}{n+\ln n} ∼ \frac{1}{n}$ ??

Answer 1

There are certainly many ways of seeing this, the easiest being probably

$$0<\frac{1}{n+\ln(n)}\leq\frac 1n\to 0$$

Answer 2

$$\frac{\frac{1}{n+\ln(n)}}{\frac{1}{n}} = \frac{1}{1+\frac{\ln{n}}{n}} \to 1 \text{ as } \frac{\ln(n)}{n} \to 0$$

Answer 3

First note that for all integers $k$ we have $$ \lim_{n \to \infty} \frac{n^k}{e^n} = 0 $$ (this can be proved using induction). Consequently, $$ \lim_{n \to \infty} \frac{\ln n}{n^\epsilon} = 0 $$ for all $\epsilon > 0$. Thus, for sufficiently large $n$ we have $\ln n \leq n^{\epsilon}$ for any $\epsilon$ (the smaller value of $\epsilon$ we take will determine exactly what we mean by "sufficiently large"). In particular, $\ln n \leq \sqrt{n}$ when $n$ is large. In this regime: $$ \frac{1}{n + \ln n} \geq \frac{1}{n + \sqrt n} = \frac{n - \sqrt{n}}{n^2 - n} \geq \frac{n - \sqrt{n}}{n^2} = \frac{1}{n} - \frac{1}{\sqrt{n}}. $$ Since $\ln n > 0$ (when $n > 1$, and thus also when $n$ is sufficiently large), we have $$ \frac{1}{n} - \frac{1}{\sqrt{n}} \leq \frac{1}{n + \ln n} \leq \frac{1}{\ln n}. $$

Answer 4

The notation ~ means that the sequences are in the same equivalence class, that is, there are constants $C_1$, $C_2$ and $N \in \mathbb{N}$ s.t. $n > N$ implies $C_1 \frac{1}{n} \leq \frac{1}{n+\log(n)} \leq C_2 \frac{1}{n}$. To show this we estimate \begin{equation} \frac{1}{2} \frac{1}{n} = \frac{1}{n+n} \leq \frac{1}{n+\log(n+1)} \leq \frac{1}{n+\log(n)} \leq 1 \frac{1}{n} \ , \end{equation} where $n > N = 0$. Hence we can choose $C_1=\frac{1}{2}$, $C_2=1$, and $N = 0$. Now the behaviour of $\frac{1}{n}$ determines the behaviour of $\frac{1}{n+\log(n)}$. In this case the sequence is convergent, but the lower limit estimate takes care of the case where a sequence have to be shown to be divergent.