To tackle this problem step-by-step, let's analyze the given sequence [tex]\( d_n = \ln \left( 4^n \right) - \ln (n!) \)[/tex]:
1. Rewrite the Sequence:
The sequence can be rewritten as:
[tex]\[
d_n = \ln \left( \frac{4^n}{n!} \right)
\][/tex]
This simplification helps in understanding the behavior of [tex]\( d_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity.
2. Exponential Function and Limit:
We can use the fact that the exponential function [tex]\( e^x \)[/tex] is continuous. Therefore, we can find the limit of [tex]\( e^{d_n} \)[/tex] and use it to find the limit of [tex]\( d_n \)[/tex]. This means:
[tex]\[
e^{d_n} = \frac{4^n}{n!}
\][/tex]
3. Evaluating [tex]\( \lim_{n \to \infty} e^{d_n} \)[/tex]:
Consider the behavior of [tex]\( \frac{4^n}{n!} \)[/tex] as [tex]\( n \)[/tex] goes to infinity. Using properties from analysis, such as Stirling's approximation, we know that:
[tex]\[
\lim_{n \to \infty} \frac{4^n}{n!} = 0
\][/tex]
This is because [tex]\( n! \)[/tex] grows much faster than [tex]\( 4^n \)[/tex].
4. Connecting [tex]\( e^{d_n} \)[/tex] and [tex]\( d_n \)[/tex]:
Since [tex]\( \lim_{n \to \infty} e^{d_n} = 0 \)[/tex], we can use the fact that if [tex]\( \lim_{n \to \infty} e^{d_n} = 0 \)[/tex], then [tex]\( \lim_{n \to \infty} d_n = -\infty \)[/tex]. The exponential function [tex]\( e^x \)[/tex] approaches 0 only when [tex]\( x \)[/tex] approaches negative infinity.
5. Conclusion:
Therefore, the limit of [tex]\( d_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity is:
[tex]\[
\lim_{n \to \infty} d_n = -\infty
\][/tex]
Since [tex]\( d_n \)[/tex] approaches negative infinity, it implies that the sequence does not converge to a finite value but rather diverges to negative infinity.
Thus, the correct conclusion for the sequence [tex]\( d_n \)[/tex] is:
[tex]\[
\lim_{n \to \infty} d_n = -\infty
\][/tex]
