Sum of increasing geometric series

Question

Sum of the series $1 + \frac{1+3}{2!}+ \frac{1+3+3^2}{3!}+....... $

The series becomes $ \sum_{k=1}^{\infty} \frac {3^{k-1}}{k!}$. How to calculate it's sum? Is it divergent due to the numerator?

Answer 1

$$b_n=\frac{1}{3}\sum_{n=1}^N3^n=\frac{1}{3}\sum_{n=0}^N3^n-\frac{1}{3}=\frac{3^{N+1}-1}{6}-\frac{1}{3}$$

So $\frac{b_N}{N!} \to 0$

Answer 2

We have series of type $\sum_{n = 1}^\infty a_n$, where $a_n = \frac{\sum_{j=1}^n 3^{j-1}}{n!} = \frac{1}{3} \frac{\sum_{j=1}^n 3^j}{n!}$

Since $\sum_{j=1}^n 3^j = 3 \cdot \frac{1 - 3^n}{1-3} = \frac{3}{2}(3^n - 1)$

So, we get $a_n = \frac{1}{2} (\frac{3^n}{n!} - \frac{1}{n!})$

And our series look like:

$\frac{1}{2} \sum_{n=1}^\infty ( \frac{3^n}{n!} - \frac{1}{n!})$

Since both $\sum_{n=1}^\infty \frac{3^n}{n!}$ and $\sum_{n=1}^\infty \frac{1}{n!} $ are convergent ( u can use D'alambert test), we can split our infinite sum into $2$ sums, getting:

$\sum_{n=1}^\infty a_n = \frac{1}{2} \sum_{n=1}^\infty \frac{3^n}{n!} - \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n!}$

Now, we should recall that $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$, so we have:

$\sum_{n=1}^\infty \frac{3^n}{n!} = e^3 - 1$ and similarly $\sum_{n=1}^\infty \frac{1}{n!} = e -1$. We finally get:

$\sum_{n=1}^\infty a_n = \frac{1}{2}(e^3 - 1 - e+ 1) = \frac{e}{2}(e^2-1)$

Answer 3

The numerator of the $k$th term: $1+3+9+\cdots+3^{k-1}=\dfrac{3^k-1}{2}$

The denominator of the $k$th term: $k!$

The sum: $$\sum_{k=1}^\infty \dfrac{3^k-1}{2\left(k!\right)}=\dfrac{1}{2}\sum_{k=1}^\infty \dfrac{3^k}{k!}-\dfrac{1}{2}\sum_{k=1}^\infty \dfrac{1}{k!}=\dfrac{1}{2}\left(e^3-1\right)-\dfrac{1}{2}\left(e-1\right)=\dfrac{e^3-e}{2}$$