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

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Answer 1 · 2018-05-22T07:21:12

# e/(e-1)#.

The sum, say #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)#