Is $\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$?

Question

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?

Answer 1

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 $1$.

Answer 2

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.

Answer 3

$$1\leq \frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}= 1+\frac{\sum_{n=1}^{k-1} 2^{2\times3^{n}}}{2^{2\times3^{k}}} = 1+ \sum_{n=1}^{k-1} 2^{-2(3^k-3^{n})} \leq 1+ \sum_{n=1}^{k-1} 2^{-2(3^k-3^{k-1})}\\=1+(k-1)2^{-4\cdot3^{k-1}} $$ 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 $$ \lim_{k\to\infty} \frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}} = 1 $$

Answer 4

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 $a > 1, b > 0, c > 1$.

Answer 5

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 $1.$

Answer 6

$$ \begin{align} \lim_{k\to\infty}\frac{\sum_{n=1}^k2^{2\cdot3^n}}{2^{2\cdot3^k}} &=\lim_{k\to\infty}\sum_{n=1}^k2^{-2\left(3^k-3^n\right)}\\ &=1+\lim_{k\to\infty}\sum_{n=1}^{k-1}2^{-2\left(3^k-3^n\right)}\\ &\le1+\lim_{k\to\infty}(k-1)2^{-2\left(3^k-3^{k-1}\right)}\\[6pt] &=1+\lim_{k\to\infty}(k-1)2^{-4\cdot3^{k-1}}\\[9pt] &=1 \end{align} $$

Answer 7

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→∞ $