Question

Evaluate $\underset{n\to \infty }{lim}\{lo{g}_{n-1}\left(n\right).{\log }_n(n+1)\dots {\log }_{n^k-1}\left(n^k\right)\}$, where $k\in N-\left\{1\right\}$.

• A. $k$
• B. $logk$
• C. $n^k$
• D. none of these

Answer 1

The correct option is A $k$

Let $P=\underset{n\to \infty }{lim}lo{g}_{n-1}\left(n\right).{\log }_n(n+1)\dots \dots \dots {\log }_{n^k-1}\left(n^k\right)$
$=\underset{n\to \infty }{lim}\{\frac{\log n}{\log (n-1)}.\frac{\log (n+1)}{\log \left(n\right)}\dots \dots \frac{\log \left(n^k\right)}{\log (n^k-1)}\}$
$=\underset{n\to \infty }{lim}\frac{\log \left(n^k\right)}{\log (n-1)}=k.\underset{n\to \infty }{lim}\frac{\log n}{\log (n-1)}=k.\underset{n\to \infty }{lim}\frac{\frac{1}{n}}{\frac{1}{n-1}}=k.\underset{n\to \infty }{lim}(1-\frac{1}{n})$
$⟹\underset{n\to \infty }{lim}\{lo{g}_{n-1}\left(n\right).{\log }_n(n+1)\dots \dots \dots {\log }_{n^k-1}\left(n^k\right)\}=k$