Is the sequence $\{(\ln (n+1)/\ln n)^n \}$ convergent ? If so then what is its limit? Is the sequence $\{(\ln (n+1)/\ln n)^n \}$ convergent ? If so then what is its limit? I tried breaking it up as 
$$\Bigg(1+ \dfrac{\ln \dfrac{n+1}n}{\ln n}\Bigg)^n,$$ but nothing's coming up. Please help.
 A: One may write
$$
\left(\frac{\ln (n+1)}{\ln n}\right)^n=\left(\frac{\ln n+\ln (1+1/n)}{\ln n}\right)^n=\left(1+\dfrac{\ln (1+1/n)}{\ln n}\right)^n
$$ then, as $n \to \infty$, use
$$
\ln \left(1+\frac1n \right)=\frac1n+O \left(\frac1{n^2} \right)
$$  giving
$$
\begin{align}
\left(\frac{\ln (n+1)}{\ln n}\right)^n&=\left(1+\frac1{n\ln n}+O \left(\frac1{n^2\ln n} \right)\right)^n
\\\\&=e^{\large n\ln \left(1+\frac1{n\ln n}+O \left(\frac1{n^2\ln n}\right)\right)}
\\\\&=e^{\large\frac1{\ln n}+O \left(\frac1{n \ln n} \right)}
\\\\& \longrightarrow 1.
\end{align}
$$
A: Clearly,
$$
a_n=\frac{\ln (n+1)}{\ln n}-1=\frac{\ln (n+1)-\ln n}{\ln n}
=\frac{\ln (1+1/n)}{\ln n}\to 0,
$$
since for $x>0$, it is clear that $0<\ln(1+x)<x$.
Thus
$$
b_n=(1+a_n)^{1/a_n}\to \mathrm{e}.
$$
Hence
$$
0<na_n=n\left(\frac{\ln (n+1)}{\ln n}-1\right)=\frac{n\big(\ln(n+1)-\ln n\big)}{\ln n}=\frac{n\ln(1+1/n)}{\ln n}<\frac{n\cdot(1/n)}{\ln n}=\frac{1}{\ln n}.
$$
Therefore 
$$
\left(\frac{\ln (n+1)}{\ln n}\right)^n=(1+a_n)^{\frac{1}{a_n}\cdot(na_n)}<(1+a_n)^{\frac{1}{a_n}\cdot\frac{1}{\ln n}}\to \mathrm{e}^{0}.
$$
