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.
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.
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,
$\displaystyle \lim_{n\to \infty} \frac{n}{\ln (1+e^n)}=1$
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!