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?

 A: Note that
$$2^{2\times3^k}\lt\sum_{n=1}^k 2^{2\times3^n}\lt2^{2\times3^k}+(k-1)2^{2\times3^{k-1}}$$
hence
$$1\lt{\sum_{n=1}^k 2^{2\times3^n}\over2^{2\times3^k}}\lt1+{k-1\over2^{2(3^k-3^{k-1})}}=1+{k-1\over16^{3^{k-1}}}\lt1+{k\over2^k}$$
where the final inequality is extremely crude, intended mainly to make it easy to see that the Squeeze Theorem tells us the limit as $k\to\infty$ is $1$.
Ah, I see José Carlos Santos had much the same idea.
A: $$1\leq \frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=
1+\frac{\sum_{n=1}^{k-1} 2^{2\times3^{n}}}{2^{2\times3^{k}}} =
1+ \sum_{n=1}^{k-1} 2^{-2(3^k-3^{n})}
\leq 1+ \sum_{n=1}^{k-1} 2^{-2(3^k-3^{k-1})}\\=1+(k-1)2^{-4\cdot3^{k-1}}
$$
So for all $k$
$$
1\leq \frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}\leq 1+(k-1)2^{-4\cdot3^{k-1}}
$$
and since 
$$
\lim_{k\to\infty}(k-1)2^{-4\cdot3^{k-1}}=0
$$
it follows that
$$
\lim_{k\to\infty} \frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}} = 1
$$
A: Is
$\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1
$?
Let
$\begin{array}\\
s(k)
&={2^{-2\times3^{k}}}\sum_{n=1}^{k} 2^{2\times3^{n}}\\
&=\sum_{n=1}^{k} 2^{2\times3^{n}-2\times3^{k}}\\
&=1+\sum_{n=1}^{k-1} 2^{2\times3^{n}-2\times3^{k}}\\
&\ge 1\\
\text{and}\\
s(k)
&\le 1+\sum_{n=1}^{k-1} 2^{2\times3^{k-1}-2\times3^{k}}\\
&= 1+(k-1)2^{2\times(3^{k-1}-3^{k})}\\
&= 1+(k-1)2^{2\times(-2\times 3^{k-1})}\\
&= 1+\dfrac{k-1}{2^{4\times 3^{k-1}}}\\
&\to 1
\qquad\text{as } k \to \infty\\
\end{array}
$
Therefore
$\lim_{k \to \infty} s(k)
= 1
$.
Note that this holds for
$\lim_{k\to\infty}\dfrac{\sum_{n=1}^{k} a^{b\times c^{n}}}{a^{b\times c^{k}}}
$
for any
$a > 1, b > 0, c > 1$.
A: Since the denominator $\to \infty,$ Stolz-Cesaro comes into play, and we consider
$$\frac{ 2^{2\cdot 3^{k+1}}}{2^{2\cdot 3^{k+1}}- 2^{2\cdot 3^{k}}} = \frac{1}{1-2^{2\cdot3^{k}-2\cdot3^{k+1}}}  = \frac{1}{1-2^{-4\cdot3^{k}}} \to 1.$$
SC thus implies the limit is $1.$ 
A: If $n<k$, then\begin{align}\frac{2^{2\times3^n}}{2^{2\times3^k}}&=4^{3^n-3^k}\\&\leqslant4^{3^{k-1}-3^k}\\&=4^{-2\times3^{k-1}}\\&\leqslant4^{-(k-1)},\end{align}because $2\times3^{k-1}\geqslant k-1$. But then$$1\leqslant\frac{\sum_{n=1}^k2^{2\times3^n}}{2^{2\times3^k}}\leqslant\frac{k-1}{4^{k-1}}+1$$So, yes, your limit is equal to $1$.
A: $$
\begin{align}
\lim_{k\to\infty}\frac{\sum_{n=1}^k2^{2\cdot3^n}}{2^{2\cdot3^k}}
&=\lim_{k\to\infty}\sum_{n=1}^k2^{-2\left(3^k-3^n\right)}\\
&=1+\lim_{k\to\infty}\sum_{n=1}^{k-1}2^{-2\left(3^k-3^n\right)}\\
&\le1+\lim_{k\to\infty}(k-1)2^{-2\left(3^k-3^{k-1}\right)}\\[6pt]
&=1+\lim_{k\to\infty}(k-1)2^{-4\cdot3^{k-1}}\\[9pt]
&=1
\end{align}
$$
A: 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→∞ $
