Find the smallest value of N that will make the inequality hold for all n N.
-$$a _ { n } = ( 1 + 0.153 / n ) ^ { n }$$
A) L ≈ 1.1653, N = 6
B) L ≈ 1.153, N = 6
C) L ≈ 1.1653, N = 2
D) diverges

Question

Find the smallest value of N that will make the inequality hold for all n > N.
-$$a _ { n } = ( 1 + 0.153 / n ) ^ { n }$$
A) L ≈  1.1653, N = 6
B) L ≈  1.153, N = 6
C) L ≈  1.1653, N = 2
D) diverges

Quizplus · Accepted Answer

First, we take the limit of the sequence as n approaches infinity:
\begin{align*}
\lim_{n 	o \infty} a_n &= \lim_{n 	o \infty} \left(1+\frac{0.153}{n}ight)^n \
&= e^{0.153} \
&\approx 1.1653 \
\end{align*}
Now, we need to find the smallest value of N such that $a_n \geq 1.1653$ for all $n > N$. We can rearrange the inequality as follows:
\begin{align*}
a_n &\geq 1.1653 \
\left(1+\frac{0.153}{n}ight)^n &\geq 1.1653 \
n\ln\left(1+\frac{0.153}{n}ight) &\geq \ln(1.1653) \
\ln\left(1+\frac{0.153}{n}ight) &\geq \frac{\ln(1.1653)}{n} \
\end{align*}
Let $f(x) = \ln\left(1+\frac{0.153}{x}ight)$. Then $f'(x) = -\frac{0.153}{x(x+0.153)} < 0$ for all $x > 0$. Therefore, $f(x)$ is a decreasing function. 
If we want $\ln\left(1+\frac{0.153}{n}ight) \geq \frac{\ln(1.1653)}{n}$ for all $n > N$, then we need $f(N) \geq \frac{\ln(1.1653)}{N}$. Rearranging, we get $\frac{0.153}{N+0.153} \geq \frac{\ln(1.1653)}{N}$. It turns out that $N = 2$ satisfies this inequality, but we need to check that it works for all $n > 2$:
\begin{align*}
\ln\left(1+\frac{0.153}{n}ight) &\geq \frac{\ln(1.1653)}{n} \
\frac{0.153}{n+0.153} &\geq \frac{\ln(1.1653)}{n} \
n^2 &\geq 0.153\ln(1.1653)n + 0.153^2 \
n &\geq 5.284 \
\end{align*}
Therefore, $N=2$ works for all $n > 5$. However, we need to choose the smallest possible value of $N$, so the answer is $\boxed{	extbf{(C) }L \approx 1.1653, N=2}$.

Find the Smallest Value of N That Will Make the Inequality