Determine the sum of the series 1, 1/2, 1/4, 1/8 .......... t(14)?

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Answer 1 · 2016-06-07T15:49:55

#sum_{i=0}^oo 1/2^i = 2# or
# sum_{i=0}^14 (1/2)^i ==32767/16384 = 1.9999389648437500000#

#sum_{i=0}^oo 1/2^i = sum_{i=0}^oo (1/2)^i#

We know that

#(1-x^{n+1})/(1-x) = 1 + x + x^2+ cdots + x^n# and that

for #abs(x)<1# we have#lim_{n->oo}(1-x^{n+1})/(1-x)=1/(1-x)#

Now #1/2 < 1# so

#sum_{i=0}^oo (1/2)^i = 1/(1-1/2) = 2#

also if the sumation is done from #i=0# to #i = 14#
then

# sum_{i=0}^14 (1/2)^i = (1-(1/2)^{15+1})/(1-(1/2)) =32767/16384 = 1.9999389648437500000#