1. ## Logarithmic limit?

Evaluate:

$\lim_{n\rightarrow \infty} \{\log_{n - 1} n\cdot \log_{n} (n + 1)\cdot \log_{n + 1} (n + 2)\mbox{...}\log_{n^k - 1} n^k\}$, where $k\in\mathbb{N}$

2. $\log_{n-1}(n)\cdot\log_{n}(n+1)\cdot\log_{n+1}(n+2)\cdot.. .\cdot\log_{n^k-1}(n^k)=$ $\frac{\ln(n)}{\ln(n-1)}\frac{\ln(n+1)}{\ln(n)}\frac{\ln(n+2)}{\ln(n+1) }\cdot ... \cdot \frac{\ln(n^k)}{\ln(n^k-1)}=$ $\frac{\ln(n^k)}{\ln(n-1)}$

Using L'Hospital's rule I get the answer to be $k$.

3. Brilliant work, spec!

I'd finish it this way . . .

$\lim_{n\to\infty}\frac{\log(n^k)}{\log(n-1)} \:=\:\lim_{n\to\infty}\frac{k\!\cdot\!\log(n)}{\lo g(n-1)}\;=\;k\cdot\lim_{n\to\infty}\frac{\log(n)}{\log (n-1)} \;=\;k\cdot1 \:=\:k$