# Limit Of Sequence

• Jan 21st 2008, 02:23 PM
Limit Of Sequence
Does someone mind showing me how to compute this limit algebraically? I'm pretty sure it goes to e, but I can't figure how to show it. Thanks for your help.

lim (n+1)^(1/ln(n+1))
• Jan 21st 2008, 03:11 PM
Krizalid
Quote:

Does someone mind showing me how to compute this limit algebraically? I'm pretty sure it goes to e, but I can't figure how to show it. Thanks for your help.

lim (n+1)^(1/ln(n+1))

$\displaystyle \mathop {\lim }\limits_{n \to \infty } (n + 1)^{1/\ln (n + 1)}.$

Substitute $\displaystyle u = \frac{1} {{\ln (n + 1)}},$ the limit becomes

$\displaystyle \mathop {\lim }\limits_{u \to 0} \left( {e^{1/u} } \right)^u = \mathop {\lim }\limits_{u \to 0} e.$

The rest follows.
• Jan 21st 2008, 04:05 PM
clearly $\displaystyle \left( e^{1/u} \right)^u = e$, so the result follows if we take the limit as u goes to zero. i suppose you are wondering where $\displaystyle e^{1/u}$ came from?
note that $\displaystyle e^{\ln x} = x$
thus: $\displaystyle n + 1 = e^{\ln (n + 1)}$, but since $\displaystyle u = \frac 1{\ln (n + 1)}$, we have that $\displaystyle \ln (n + 1) = \frac 1u$, thus it follows that $\displaystyle e^{\ln (n + 1)} = e^{1/u}$