What is the sum of the infinite geometric series $\infty \sum n = 0 {(\frac{1}{e})}^{n}$ $sum_(n=0)^oo(1/e)^n$ ?

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Answer 1 · 2018-05-22T07:21:12

$\frac{e}{e - 1}$ $e/(e-1)$ .

The sum, say $s$ $s$ , of the infinite Geometric Series :

$a+ar+ar^2+...+ar^(n-1)+...$ is given by,

$s=a/(1-r), if |r| lt 1$ .

We are given the infinite geometric series :

$1+1/e+1/e^2+...$ .

$:. a=1, r=1/e," so that, "|r|=|1/e| lt 1$ .

Hence, $s=1/{1-(1/e)}=e/(e-1)$ .

Answer 2 · 2018-05-22T07:25:22

$sum_(n=0)^oo (1/e)^n = e/(e-1)$

The general term of a geometric series can be written:

$a_k = a r^(k-1)" "$ ( $k = 1,2,3,...$ )

where $a$ is the initial term and $r$ the common ratio.

If $abs(r) < 1$ then the series converges with sum:

$s_oo = a/(1-r)$

In our example $a = (1/e)^0 = 1$ and $r = 1/e$

so the sum is:

$sum_(n=0)^oo (1/e)^n = 1/(1-1/e) = e/(e-1)$

What is the sum of the infinite geometric series sum_(n=0)^oo(1/e)^n∞∑n=0(1e)nsum_(n=0)^oo(1/e)^n ?