How do you apply the ratio test to determine if #Sigma 2^n/(n!)# from #n=[1,oo)# is convergent to divergent?

Question

Christopher Agresta · Accepted Answer

The series is convergent and:

#sum_(n=1)^oo 2^n/(n!) = e^2-1# The ratio test states that a sufficient condition for a series: #sum_(n=0)^oo a_n# to converge absolutely is that: #L = lim_(n->oo) abs(a_(n+1)/a_n) < 1# We then evaluate the ratio for the series: #sum_(n=1)^oo 2^n/(n!)# #abs(a_(n+1)/a_n) = (2^(n+1)/((n+1)!))/(2^n/(n!)) = 2^(n+1)/2^n (n!)/((n+1)!) = 2/(n+1)# so that: #lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) 2/(n+1) = 0# proving that the series is convergent. We should note by the way that the sum is: #sum_(n=1)^oo 2^n/(n!) = -1 + [sum_(n=0)^oo x^n/(n!)]_(x=2) = e^2-1#

Emilia Aguilar · Answer

undefined To apply the ratio test to determine the convergence or divergence of the series $\sum_{n=1}^{\infty} \frac{2^n}{n!}$, we evaluate the limit of the absolute value of the ratio of successive terms as $n$ approaches infinity:

$$
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
$$

where $a_n = \frac{2^n}{n!}$.

First, we find $a_{n+1}$ and $a_n$:

$$
a_{n+1} = \frac{2^{n+1}}{(n+1)!} = \frac{2 \cdot 2^n}{(n+1)n!} = \frac{2}{n+1} \cdot \frac{2^n}{n!}
$$

$$
a_n = \frac{2^n}{n!}
$$

Next, we take the ratio of these terms:

$$
\frac{a_{n+1}}{a_n} = \frac{\frac{2}{n+1} \cdot \frac{2^n}{n!}}{\frac{2^n}{n!}} = \frac{2}{n+1}
$$

Finally, we find the limit of this ratio as $n$ approaches infinity:

$$
L = \lim_{n \to \infty} \frac{2}{n+1} = 0
$$

Since the limit of the ratio is less than 1, by the ratio test, the series $\sum_{n=1}^{\infty} \frac{2^n}{n!}$ converges.