Convergence $\left(1+ \frac{\ln (1+1/n)^n}{\ln n^n}\right)^{\ln n^n} \to e$

Question

Convergence of sequence whose $n^{th}$ term is $$\left(1+ \frac{\ln (1+\frac{1}{n})^n}{\ln n^n}\right)^{\ln n^n} \to e \\ T_n = e^{{\ln n^n}\ln\left(1+ \frac{\ln (1+\frac{1}{n})^n}{\ln n^n}\right)}$$

After that what should i do? It is mentioned in To test the convergence of a series whose nth term is $a_{n}=\left({\frac{\log n}{log(n+1)}}\right)^{n^2\log n}$ using Root test But i did not understand part after "indeed" by 2nd Solution. Please help me in understanding it:

$(\dots)$ indeed $$\left(1+\frac{\log(1+\frac1n)^n}{\log n^n}\right)^{\log n^n}=e^{\log n^n\cdot \log \left(1+\frac{\log(1+\frac1n)^n}{\log n^n}\right)}=e^{\log n^n\cdot \left(\frac{\log(1+\frac1n)^n}{\log n^n}+o\left(\frac{1}{logn^n}\right)\right)}=e^{\left(\log(1+\frac1n)^n+o(1)\right)}\to e$$

Or is there any alternate way?

Answer 1

We have that

$$\frac{\ln \left(1+\frac{1}{n}\right)^n}{\ln n^n} \to 0$$

indeed

• $\ln \left(1+\frac{1}{n}\right)^n\to \ln e=1$
• $\ln n^n\to \infty$

therefore

$$\left(1+ \frac{\ln (1+\frac{1}{n})^n}{\ln n^n}\right)^{\ln n^n}=\left[\left(1+ \frac{\ln (1+\frac{1}{n})^n}{\ln n^n}\right)^{\frac{\ln n^n}{\ln (1+\frac{1}{n})^n}}\right]^{\ln (1+\frac{1}{n})^n} \to e^1=e$$

indeed let $m=\frac{\ln \left(1+\frac{1}{n}\right)^n}{\ln n^n} \to 0$ and thus

$$\left(1+ \frac{\ln (1+\frac{1}{n})^n}{\ln n^n}\right)^{\frac{\ln n^n}{\ln (1+\frac{1}{n})^n}}=(1+m)^\frac1m \to e$$