Question

Determine whether the series is convergent or divergent. If its convergent, find its sum.

\(\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{(n - 1)}}}}} \)

Short answer

The given geometric series isdivergent.

Step-by-step solution

Rewriting the equation.

\(\begin{aligned}{l}\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {{3^n}{e^{{{(n - 1)}^{ - 1}}}}} \\\end{aligned}\)

\( = \sum\limits_{n = 1}^\infty {{3^n}{e^{(1 - n)}}} \)

Writing in terms of b and q.

\(\sum\limits_{n = 1}^\infty {{3^n}{e^{(1 - n)}}} \)is a geometric series with first term b = \(\frac{3}{e}\)and common ratio q = \(\frac{3}{e}\).

Applying ratio test.

Since \(\left| q \right| = \left| {\frac{3}{e}} \right| > 1\), then by ratio test the series is divergent.

Hence, \(\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{n - 1}}}}} \) is divergent.