Show that #3^(1/3)xx9^(1/9)xx27^(1/27)...# to infinity #=3^(3/4)#.how?

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Answer 1 · 2018-03-03T17:51:04

See below.

#3^(1/3)xx9^(1/9)xx27^(1/27)cdots =3^(1/3) xx 3^(2/9) xx 3^(3/27) cdots = 3^(1/3+2/9+3/27+ cdots+ n/3^n+ cdots) = 3^S#

with

#S = sum_(k=1)^oo n/3^n = ?#

We know that #sum_(k=1)^oo k x^k = x d/(dx) sum_(k=1)^oo x^k#

and also that for #abs x < 1#

#sum_(k=1)^oo x^k = 1/(1-x)-1# and #d/(dx) (1/(1-x)-1) = 1/(1-x)^2# then

#sum_(k=1)^oo k x^k = x/(1-x)^2# and for #x = 1/3# we have

#S = 3/4# then finally

#3^S = 3 ^(3/4)#