\sum _{n=3\:}^{\infty }\frac{e^{4n}}{(n-2)!} convergence or divergence

asked by guest
on Oct 07, 2024 at 7:59 pm



You asked:

Determine whether the following series converges or diverges: \(\displaystyle\sum_{n = 3}^{\infty} {\frac{{e}^{4 n}}{\left( n - 2 \right)!}}\)

MathBot Answer:

The infinite series \(\displaystyle\sum_{n=3}^{\infty} \frac{e^{4 n}}{\left(n - 2\right)!}\) converges.

We can use the ratio test to determine convergence of this series.

Given a series \(\displaystyle\sum a_n\), define \(L = \displaystyle\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|\)
 •If \(L < 1\) the series converges absolutely.
 •If \(L > 1\) the series diverges.
 •If \(L = 1\) or the limit does not exist the test is inconclusive.
$$\begin{aligned}L &= \lim_{n \to \infty}\left|\frac{\frac{e^{4 n + 4}}{\left(n - 1\right)!}}{\frac{e^{4 n}}{\left(n - 2\right)!}}\right| \\ &= \lim_{n \to \infty} e^{4} \left|{\frac{\left(n - 2\right)!}{\left(n - 1\right)!}}\right| \\ &= 0\end{aligned}$$Therefore the series converges.

\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).

+ Submit Question