Determine if the series converges or diverges; if the series converges, find its sum.
-$$\sum _ { n = 0 } ^ { \infty } e ^ { - 4 n }$$
A) Converges; $\frac { e ^ { 4 } } { e ^ { 4 } - 1 }$
B) Converges; $\frac { 1 } { \mathrm { e } ^ { - 4 } - 1 }$
C) Converges; $\frac { e ^ { - 4 } } { e ^ { - 4 } - 1 }$
D) Diverges

Question

Quizplus · Accepted Answer

This series is a geometric series with the first term $a = e^0 = 1$ and common ratio $r = e^{-4}$. A geometric series converges if $|r| < 1$, which is true in this case. The sum of a convergent geometric series is given by $S = \frac{a}{1 - r}$, so the sum is $S = \frac{1}{1 - e^{-4}} = \frac{e^4}{e^4 - 1}$.

Determine If the Series Converges or Diverges; If the Series ∑n=0∞e−4n\sum _ { n = 0 } ^ { \infty } e ^ { - 4 n }n=0∑∞​e−4n

Determine If the Series Converges or Diverges; If the Series $\sum _ { n = 0 } ^ { \infty } e ^ { - 4 n }$