Evaluate $3\times(9^{(\frac12)})\times(27^{(\frac14)})*(81^{(\frac18)}\cdots\infty$

Question

Here expression can be written as $$3\times(3^{(\frac22)})\times(3^{(\frac34)})\times(3^{(\frac48)}).... = 3^{(1+1+(\frac34)+(\frac48)+(\frac5{16})\cdots\infty)}$$

What will be the next step after this?

Answer 1

So your sum is

$$ 3^{\sum_{k=0}^\infty (k+1)/2^k} $$

The exponent can be evaluated: $$ \sum_{k=0}^\infty (k+1)/2^k = \sum_{k=0}^\infty 1/2^k + \sum_{k=0}^\infty k/2^k = 2 + 2 = 4 $$

So your sum is $3^4 = 81$.

EDIT by request: The last sum can be evaluated as follows: $$ \frac{\partial}{\partial x} \sum_{k=0}^\infty x^{-k} = -\sum_{k=0}^\infty k/x^{k+1} = - \frac1x \sum_{k=0}^\infty k/x^{k} $$ but also $$ \frac{\partial}{\partial x} \sum_{k=0}^\infty x^{-k} = \frac{\partial}{\partial x} \frac{1}{1-1/x} = -1/(x - 1)^2 $$ Equating the two results gives $$ \sum_{k=0}^\infty k/x^{k} = x/(x - 1)^2 $$ For $x=2$ we obtain $$ \sum_{k=0}^\infty k/2^{k} = 2 $$

Answer 2

Here is the way to find the exponent Let S $$S=\sum_{k=0}^\infty(\frac{k+1}{2^k})=(1+\frac22+\frac34+...)$$ $$2S=2+(1+\frac32+\frac44+...)$$ Subtracting gives $$S=2+(1+\frac32+\frac44+...)-(1+\frac22+\frac34+...)$$ Note that we can rearrange as the series converges $$S=2+(1+\frac12+\frac14+...)$$ The terms in bracket is a Geometric infinite series $$S=2+(\frac{1}{1-\frac12})$$ $$S=4$$ So $$3^{\sum_{k=0}^\infty(\frac{k+1}{2^k})}=3^{4}=81$$