# find the limit of the given sequence.

• Jan 30th 2007, 03:44 PM
rcmango
find the limit of the given sequence.
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.
• Jan 30th 2007, 05:08 PM
ThePerfectHacker
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,
$\lim_{n\to \infty} \frac{n}{\ln (1+e^n)}=1$
• Jan 30th 2007, 07:29 PM
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!
• Jan 30th 2007, 07:48 PM
ThePerfectHacker
Quote:

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,
$1+e^n\leq e^n+e^n=2e^n$
• Jan 30th 2007, 10:15 PM
rcmango
okay, i can see as both sides of the squeeze theorem converge to 1.
then, also the middle equation converges to 1.