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))

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- Jan 21st 2008, 02:23 PMjamesHADDYLimit 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 PMKrizalid
$\displaystyle \mathop {\lim }\limits_{n \to \infty } (n + 1)^{1/\ln (n + 1)}.$

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 PMjamesHADDY
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 PMJhevon
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 $\displaystyle e^{\ln (n + 1)} = e^{1/u}$