## Related questions with answers

Find the sum of the series. 3+9/2!+27/3!+81/4!+. . .

Solutions

Verified$3 + \dfrac{9}{2!} + \dfrac{27}{3!} + \dfrac{81}{4!} + \cdots =\dfrac{3^1}{1!} + \dfrac{3^2}{2!} + \dfrac{3^3}{3!} + \dfrac{3^4}{4!} + \cdots = \sum_{n=1}^{\infty} \dfrac{3^n}{n!}$

First we write the series as a summation.

$\sum_{n=1}^{\infty}\dfrac{3^n}{n!} = \dfrac{3^1}{1!}+\dfrac{3^2}{2!}+\dfrac{3^3}{3!}+\dfrac{3^4}{4!}+\dots$

NOTE: our index starts at $n=1$ since $n=0$ will give us 1, which is not a value in our series.

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 $3$ - therefore, the general term will be $3^n$. Then, notice that the denominators are factorials. That's why the denominator in the general term will be $n!$, and we have the series:

$\begin{align*} 3 + \frac{9}{2!} + \frac{27}{3!} + \frac{81}{4!} + ... = \sum_{n=1}^\infty \frac{3^n}{n!} \end{align*}$

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 $e^x$:

$\begin{align*} e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \end{align*}$

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

$\begin{align*} 1+ \sum_{n=1}^\infty \frac{3^n}{n!} = \sum_{n=0}^\infty \frac{3^n}{n!} = e^3 \implies \sum_{n=1}^\infty \frac{3^n}{n!} = e^3 - 1 \end{align*}$

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