We are asked to find the sum of the infinite series $\sum_{k=1}^{\infty} (\frac{e}{\pi})^{k+1}$.

1. Problem Description

We are asked to find the sum of the infinite series $\sum_{k=1}^{\infty} (\frac{e}{\pi})^{k+1}$.

2. Solution Steps

The given series is a geometric series. A geometric series has the form $\sum_{k=0}^{\infty} ar^k$, where $a$ is the first term and $r$ is the common ratio. The sum of an infinite geometric series converges to $\frac{a}{1-r}$ if $|r| < 1$.

In this case, we have $\sum_{k=1}^{\infty} (\frac{e}{\pi})^{k+1}$.

We can rewrite this as

$\sum_{k=1}^{\infty} (\frac{e}{\pi})^{k+1} = \sum_{k=1}^{\infty} (\frac{e}{\pi}) (\frac{e}{\pi})^k = \frac{e}{\pi} \sum_{k=1}^{\infty} (\frac{e}{\pi})^k$.

Now, $\sum_{k=1}^{\infty} (\frac{e}{\pi})^k$ is a geometric series with first term $a = \frac{e}{\pi}$ and common ratio $r = \frac{e}{\pi}$.

Since $e \approx 2.718$ and $\pi \approx 3.141$, we have $\frac{e}{\pi} < 1$. Therefore, the series converges.

The sum of $\sum_{k=1}^{\infty} (\frac{e}{\pi})^k$ is $\frac{a}{1-r} = \frac{\frac{e}{\pi}}{1-\frac{e}{\pi}} = \frac{e}{\pi - e}$.

So, we have

$\sum_{k=1}^{\infty} (\frac{e}{\pi})^{k+1} = \frac{e}{\pi} \sum_{k=1}^{\infty} (\frac{e}{\pi})^k = \frac{e}{\pi} \cdot \frac{\frac{e}{\pi}}{1 - \frac{e}{\pi}} = \frac{e}{\pi} \cdot \frac{e}{\pi - e} = \frac{e^2}{\pi(\pi - e)}$.

Alternatively, we can write the original series as $\sum_{k=1}^\infty (\frac{e}{\pi})^{k+1} = \sum_{k=2}^\infty (\frac{e}{\pi})^k$.

The series $\sum_{k=0}^\infty (\frac{e}{\pi})^k = \frac{1}{1 - \frac{e}{\pi}} = \frac{\pi}{\pi - e}$.

Then $\sum_{k=2}^\infty (\frac{e}{\pi})^k = \sum_{k=0}^\infty (\frac{e}{\pi})^k - 1 - \frac{e}{\pi} = \frac{\pi}{\pi - e} - 1 - \frac{e}{\pi} = \frac{\pi - (\pi - e) - \frac{e}{\pi}(\pi - e)}{\pi - e} = \frac{e - e + \frac{e^2}{\pi}}{\pi - e} = \frac{\frac{e^2}{\pi}}{\pi - e} = \frac{e^2}{\pi(\pi - e)}$.

3. Final Answer

$\frac{e^2}{\pi(\pi - e)}$