<s>[INST] Is $\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$? Look at this limit. I think, this equality is true.But I'm not sure.
$$\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$$
For example, $k=3$, the ratio is $1.000000000014$

Is this limit mathematically correct?

 [/INST] I have taken a simpler approach to solve this.
$$\frac{\sum_{n=1}^k 2^{2X3^n}}{2^{2X3^k}}$$ can be written as:
$$\frac{\sum_{n=1}^k 4^{3^n}}{4^{3^k}}$$ which further can be written as :
$$\frac{4^{3^1}}{4^{3^k}} + \frac{4^{3^2}}{4^{3^k}} +\frac{4^{3^3}}{4^{3^k}}.........  \frac{4^{3^{k-1}}}{4^{3^k}} +\frac{4^{3^k}}{4^{3^k}}$$
writing this in reverse order:
$$\frac{4^{3^k}}{4^{3^k}} + \frac{4^{3^{k-1}}}{4^{3^k}}......... +\frac{4^{3^3}}{4^{3^k}} + \frac{4^{3^2}}{4^{3^k}} + \frac{4^{3^1}}{4^{3^k}}$$
which becomes 
$$\frac{1}{1} + \frac{1}{4^{3^k-3^{k-1}}}......... +\frac{1}{4^{3^k-3^3}} + \frac{1}{4^{3^k-3^2}} + \frac{1}{4^{3^k-3^1}}$$
Now as $k→∞, 3^k →∞ $
which implies, $4^{3^k-3^{k-1}} →∞ $
which imply, $\frac{1}{4^{3^k-3^{k-1}}} + ......... → 0 $
which leaves, $\frac{1}{1} + \frac{1}{4^{3^k-3^{k-1}}}......... +\frac{1}{4^{3^k-3^3}} + \frac{1}{4^{3^k-3^2}} + \frac{1}{4^{3^k-3^1}} → 1 $ as $k→∞ $
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