$2^{1/4} \times 4^{1/8} \times 8^{1/16} \times 16^{1/32} \times \ldots\to2$ $2^{1/4} \times 4^{1/8} \times 8^{1/16} \times 16^{1/32} \times \ldots\to2$
How can I explain this to a school student who doesn't know what a  limit is? 
 A: $\displaystyle4^{\dfrac18}=(2^2)^{\dfrac18}=2^{\dfrac28}$
$\displaystyle8^{\dfrac1{16}}=(2^3)^{\dfrac1{16}}=2^{\dfrac3{16}}$
$\displaystyle16^{\dfrac1{32}}=(2^4)^{\dfrac1{32}}=2^{\dfrac4{32}}$
So, the required limit is $$2^{\left(\dfrac14+\dfrac28+\dfrac3{16}+\dfrac4{32}+\cdots\right)}$$
$\displaystyle\dfrac14+\dfrac28+\dfrac3{16}+\dfrac4{32}+\cdots=\sum_{r=1}^\infty\frac r{2^{r+1}}$ which is an Arithmetico-geometric sequence
See also : How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$
A: $$2^{1/4} \times 4^{1/8} \times 8^{1/16} \times 16^{1/32} \times \ldots=\prod^{\infty}_{n=1}(2^n)^{1/2^{(n+1)}}=x$$
Log by base 2 both sides and we have:
$$1/4+1/4+3/16+1/8+5/64..=\sum^{\infty}_{n=1} n/2^{(n+1)}$$
This sum is $1$. To show it we need to find 
$$S=\lim \sum^{\infty}_{n=1} n/2^{(n+1)}=1$$
Let's multiply it by $2$ and substitute $S_0=2S$
$$1/2+2/4+3/8+4/16...=S_0=\lim \sum^{\infty}_{n=1} n/2^{n}=2$$
Now let's take 
$$f(x)=\frac{1}{1-x} =1+x+x^2+x^3... $$
$$f'(x)=\frac{1}{(1-x)^2}=1+2*x+3*x^2+4*x^3...$$
Multiply it by $x$
$$g(x)=f'(x)*x=\frac{x}{(1-x)^2}=1*x+2*x^2+3*x^3+4*x^4...$$
And finally the magic, watching $g(1/2)$
$$S_0=g(1/2)=2=1/2+2/4+3/8+4/16...$$
Going back from $S_0$ to $S$ and from $S$ to logarithms solving $\log_2 x = S=1 $and get final answer $$x =\prod^{\infty}_{n=1}(2^n)^{1/2^{(n+1)}}=2$$
No ideas how to show it without Taylor series and derivative.
