Finding summation for a certain $n$ to an equation involving $e$ and $\pi$, summation of $(\pi^n)/(e^n)$ from $n= 0 \to \infty$ at $n=100$?

Question

How do you find the summation of n = 100 (or any other value for that matter) for equations like:

$\sum_{n=0}^{\infty } \frac{\pi^{n}}{e^{n}}$

and

$\sum_{n=1}^{\infty } \frac{e^{n}}{\pi^{n}}$ ?

Answer 1

Since $\pi > e$, the first series diverges.

For the second one, $$\sum\limits_{n=1}^{\infty} \frac{e^n}{\pi^n} = \frac{e}{\pi}\cdot\sum\limits_{n=0}^{\infty} \frac{e^n}{\pi^n} = \frac{e}{\pi} \cdot\frac{1}{1 - \frac{e}{\pi}} = \frac{e}{\pi - e}$$

Once you have this nice equation in hand, it is straight-forward to establish $$\sum\limits_{n=0}^{N} \frac{e^n}{\pi^n} = \sum\limits_{n=0}^{\infty} \frac{e^n}{\pi^n} - \sum\limits_{n=N+1}^{\infty} \frac{e^n}{\pi^n} = \sum\limits_{n=0}^{\infty} \frac{e^n}{\pi^n} - \left(\frac{e}{\pi}\right)^{N+1}\cdot \sum\limits_{n=0}^{\infty} \frac{e^n}{\pi^n}$$

In other words $$\sum\limits_{n=0}^{N} \frac{e^n}{\pi^n} = \frac{1}{1-\frac{e}{\pi}}-\left(\frac{e}{\pi}\right)^{N+1}\cdot \frac{1}{1-\frac{e}{\pi}} = \left(1 - \left(\frac{e}{\pi}\right)^{N+1}\right)\cdot \frac{\pi}{\pi-e}$$

Answer 2

Let us assume that you are asking to compute the partial sum

$$S_n:=\sum_{k=0}^n\frac{\pi^k}{e^k}$$ and let us take the case of $n=3$.

$$S_3=1+\frac\pi e+\frac{\pi^2}{e^2}+\frac{\pi^3}{e^3}.$$

If we multiply that by $\dfrac\pi e$, we get

$$\frac\pi eS_3=\frac\pi e+\frac{\pi^2}{e^2}+\frac{\pi^3}{e^3}+\frac{\pi^4}{e^4},$$ which we can interpret as

$$\frac\pi eS_3=\color{green}{-1}+1+\frac\pi e+\frac{\pi^2}{e^2}+\frac{\pi^3}{e^3}+\color{green}{\frac{\pi^4}{e^3}}=\color{green}{-1}+S_3+\color{green}{\frac{\pi^4}{e^4}}.$$

From this you can draw $S_3$, and it is a trivial matter to generalize to other $n$.

Answer 3

The first series is a geometric series $\sum_{n=0}^N r^n$ where $r$ is the ratio between successive terms. In this case $r=\frac \pi e$.

Let's call the sum of the first $N$ terms $s(N)$. So

$s(N) = \sum_{n=0}^N r^n \\ \Rightarrow rs(N) = \sum_{n=0}^N r^{n+1} = \sum_{n=1}^{N+1} r^n \\ \Rightarrow rs(N) = s(N) + r^{N+1} - 1\\ \Rightarrow s(N) = \frac{r^{N+1}-1}{r-1}$

Set $r=\frac \pi e$ and $N=100$ and you can find the value of $s(100)$.

The second series is similar, except that $r= \frac e \pi$ and the series starts at $n=1$ instead of at $n=0$, so you have to subtract $1$ from the formula given above for $s(N)$ to get the right answer.