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

Question

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

Scarlett Allison · Accepted Answer

# 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)#.

Emily Ammerman · Answer

#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)#

Evelyn Agard · Answer

undefined The sum of the infinite geometric series $ \sum_{n=0}^{\infty} \left(\frac{1}{e}\right)^n $ can be found using the formula for the sum of an infinite geometric series, which is given by:

$$ S = \frac{a}{1 - r} $$

Where $ a $ is the first term of the series and $ r $ is the common ratio. In this case, the first term $ a $ is $ 1 $ (since $ n = 0 $) and the common ratio $ r $ is $ \frac{1}{e} $.

Substitute these values into the formula:

$$ S = \frac{1}{1 - \frac{1}{e}} $$

Simplify the expression:

$$ S = \frac{1}{1 - \frac{1}{e}} = \frac{1}{\frac{e - 1}{e}} = \frac{e}{e - 1} $$

So, the sum of the infinite geometric series $ \sum_{n=0}^{\infty} \left(\frac{1}{e}\right)^n $ is $ \frac{e}{e - 1} $.