How do I calculate this summation: $\sum_{n=1}^{k} 2^{2\times3^{n}}$

Question

How do I calculate this summation? $$\sum_{n=1}^{k} 2^{2\times3^{n}}$$

I couldn't find a way that works.

Edit:

Fundamentally to Yves Daoust's comment is, can we write ?

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

Answer 1

This is probablky too long for a comment.

As Gerry Myerson commented, I do not think that there is any closed form for this summation (even using special functions).

However, computing the first terms (up to $k=18$ for which $S_{18}\approx 1.769\times 10^{233250376}$ and plotting the results, it seems that $$\log (\log (S_k))=a + b\,k$$ could be a rather good approximation. A linear regression leads to $a=0.326638$ and $b=1.098610$. Using it for $k=10$, this would lead to $$\log (\log (S_{10}))=11.3128\implies S_{10}=1.185\times 10^{35551}$$ while the exact value is $\approx 1.097\times 10^{35551}$.

These are really very big numbers.

Edit

It seems to be nicer using $$\log_2 (\log_2 (S_k))=1+1.58496\, k$$

Answer 2

With an excellent approximation,

$$\sum_{n=1}^k 2^{2\cdot3^n}\approx2^{2\cdot3^k}.$$

For example, already with $k=3$,

$$2^6+2^{18}+2^{54}=64+262144+18014398509481984\approx 18014398509481984.$$

The ratio is $1.000000000014\cdots$