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.

Printable View

- Jan 30th 2007, 02:44 PMrcmangofind 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, 04:08 PMThePerfectHacker
Observe the inequalities,

$\displaystyle \ln (1+e^n)\geq \ln (e^n)=n$

$\displaystyle \ln (1+e^n)\leq \ln (e^n+e^n)=\ln (2e^n)=\ln 2+\ln e^n=n+\ln 2$

Thus,

$\displaystyle \frac{1}{\ln (1+e^n)} \leq \frac{1}{n}$

$\displaystyle \frac{1}{\ln(1+e^n)} \geq \frac{1}{n+\ln 2}$

Thus, multiply by $\displaystyle n>0$,

$\displaystyle \frac{n}{n+\ln 2}\leq \frac{n}{\ln (1+e^n)}\leq \frac{n}{n}=1$

Now,

$\displaystyle \lim_{n\to \infty} \frac{n}{n+\ln 2}=1$

$\displaystyle \lim_{n\to \infty} 1=1$

Thus, by the squeeze theorem,

$\displaystyle \lim_{n\to \infty} \frac{n}{\ln (1+e^n)}=1$ - Jan 30th 2007, 06:29 PMrcmango
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, 06:48 PMThePerfectHacker
- Jan 30th 2007, 09:15 PMrcmango
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.