find the limit of the given sequence.

**rcmango** · January 30th 2007, 02:44 PM

If the limit exists, find the limit of the sequence:

heres the equation: http://img402.imageshack.us/img402/5907/untitledav2.jpg

any help with this one please.

**ThePerfectHacker** · January 30th 2007, 04:08 PM

Observe the inequalities,
$\ln (1+e^n)\geq \ln (e^n)=n$
$\ln (1+e^n)\leq \ln (e^n+e^n)=\ln (2e^n)=\ln 2+\ln e^n=n+\ln 2$

Thus,
$\frac{1}{\ln (1+e^n)} \leq \frac{1}{n}$
$\frac{1}{\ln(1+e^n)} \geq \frac{1}{n+\ln 2}$
Thus, multiply by $n>0$ ,
$\frac{n}{n+\ln 2}\leq \frac{n}{\ln (1+e^n)}\leq \frac{n}{n}=1$

Now,
$\lim_{n\to \infty} \frac{n}{n+\ln 2}=1$
$\lim_{n\to \infty} 1=1$
Thus, by the squeeze theorem,
$\lim_{n\to \infty} \frac{n}{\ln (1+e^n)}=1$

**rcmango** · January 30th 2007, 06:29 PM

okay, i see that we used the squeeze theorem, however,

in the second part of 'Observe the inequalities' where it = n + ln2

why did we raise 1 to e^n?

also, we used 1/n for a bound in the squeeze theorem, is this because you recognized it as a series? if so, which one.

thanks for all the help!

**ThePerfectHacker** · January 30th 2007, 06:48 PM

Originally Posted by rcmango

okay, i see that we used the squeeze theorem, however,

in the second part of 'Observe the inequalities' where it = n + ln2

why did we raise 1 to e^n?

also, we used 1/n for a bound in the squeeze theorem, is this because you recognized it as a series? if so, which one.

thanks for all the help!

Because,
$1\leq e^n$
Thus,
$1+e^n\leq e^n+e^n=2e^n$

**rcmango** · January 30th 2007, 09:15 PM

okay, i can see as both sides of the squeeze theorem converge to 1.
then, also the middle equation converges to 1.

thanks for your help.

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