Question

Question: 59-64 Find the sum of the series.

62. \(\sum\limits_{n = 0}^\infty {\frac{{{3^n}}}{{{5^n}n!}}} \)

Short answer

The sum of the series is \({e^{3/5}}\)

Step-by-step solution

Step 1: The given power series

Rewrite the given power series as shown below:

Note that \({3^n}\) in the numerator and \({5^n}\) in the denominator have the same powers, so we can use the following property:

$$\(\sum\limits_{n = 0}^\infty {\frac{{{3^n}}}{{{5^n}n!}}} \)\(\frac{{{a^n}}}{{{b^n}}} = {(a/b)^n}{\rm{ }}\)\(\sum\limits_{n = 0}^\infty {\frac{{{{(3/5)}^n}}}{{n!}}} \)

Step 2: Final proof

From the table of important Maclaurin series, we have :

\(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} = {e^x}\)

Therefore

\(\begin{array}{l}\sum\limits_{n = 0}^\infty {\frac{{{{(3/5)}^n}}}{{n!}}} \\ = {e^{3/5}}\end{array}\)