# Limit Of Sequence

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• Jan 21st 2008, 02:23 PM
jamesHADDY
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:

Originally Posted by jamesHADDY
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))

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

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

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

The rest follows.
• Jan 21st 2008, 04:05 PM
jamesHADDY
Do you mind elaborating? I don't see how this substitution equals the original and how you get the limit as u goes to infinity?
• Jan 21st 2008, 05:35 PM
Jhevon
Quote:

Originally Posted by jamesHADDY
Do you mind elaborating? I don't see how this substitution equals the original and how you get the limit as u goes to infinity?

clearly $\left( e^{1/u} \right)^u = e$, so the result follows if we take the limit as u goes to zero. i suppose you are wondering where $e^{1/u}$ came from?

note that $e^{\ln x} = x$

thus: $n + 1 = e^{\ln (n + 1)}$, but since $u = \frac 1{\ln (n + 1)}$, we have that $\ln (n + 1) = \frac 1u$, thus it follows that $e^{\ln (n + 1)} = e^{1/u}$