Determine if the series converges or diverges; if the series converges, find its sum.
-$$\sum _ { n = 0 } ^ { \infty } \left( 1 + \frac { - 3 } { n } \right) ^ { 3 n }$$
A) Converges; $\mathrm { e } ^ { - 9 }$
B) Converges; $\frac { 1 } { | - 3 | - 1 }$
C) Converges; $\frac { 1 } { | - 3 | + 1 }$
D) Diverges

Question

Quizplus · Accepted Answer

The series diverges because the general term does not approach zero as $ n \to \infty $. The expression $\left(1 + \frac{-3}{n}\right)^{3n}$ approaches $e^{-9}$ as $n$ approaches infinity, but this does not satisfy the necessary condition for convergence of a series.

Determine If the Series Converges or Diverges; If the Series ∑n=0∞(1+−3n)3n\sum _ { n = 0 } ^ { \infty } \left( 1 + \frac { - 3 } { n } \right) ^ { 3 n }n=0∑∞​(1+n−3​)3n

Determine If the Series Converges or Diverges; If the Series $\sum _ { n = 0 } ^ { \infty } \left( 1 + \frac { - 3 } { n } \right) ^ { 3 n }$