Question

Find the sum of the series 63.\(3 + \frac{9}{{2!}} + \frac{{27}}{{3!}} + \frac{{81}}{{4!}} + ......\)

Short answer

\(3 + \frac{9}{{2!}} + \frac{{27}}{{3!}} + \frac{{81}}{{4!}} + ...... = {e^3} - 1\)

Step-by-step solution

Step 1: Form the series with help of examining the numerator and denominator.

In this problem we will first need to come up with a general term for the series. Notice that the numerators of all terms are just multiples of three- therefore, the general term will be \({3^n}\). Then notice that the denominator are factorials. That’s why the denominator in the general term will be \(n!\) and we have the series.

\(3 + \frac{9}{{2!}} + \frac{{27}}{{3!}} + \frac{{81}}{{4!}} + ...... = \sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{n!}}} \)

Step 2: Recognize familiar power series

Now we are trying to recognize a familiar power series similar to this one. Since there is an \(n!\) in the denominator, this ought to remind us of the power series for \({e^x}\)\({e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \)

Step 3: Examination

Notice that the counter in our sum starts at \(n = 1\), while the power series at n=0. Using this, we can now find the sum:

\(\begin{array}{l}1 + \sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{n!}}} = \sum\limits_{n = 0}^\infty {\frac{{{3^n}}}{{n!}} = {e^3}} \\\sum\limits_{n = 0}^\infty {\frac{{{3^n}}}{{n!}} = {e^3} - 1} \end{array}\)