I asked this question a few hours ago on MathStackExchange and there it received some attention but we still do not have a proof so I decided to ask it here also, in an unchanged form, and here it is:

Digit sums of numbers $3^m$ in base $10$ for $m=1,2,...,50$ are:

$3,9,9,9,9,18,18,18,27,27,27,18,27,45,36,27,27,45,36,45,27,45,54,54,63,63,81,72,72,63,81,63,72,99,81,81,90,90,81,90,99,90,108,90,99,108,126,117,108,144$.

Ratios $\dfrac {ds_{10}(3^m)}{ds_{10}(3^{m+1})}$ for $m=1,2,...,49$ to three decimal places are:

$0.333,1.000,1.000,1.000,0.500,1.000,1.000,0.666,1.000,1.000,1.500,0.666,0.600,1.250,1.333,1.000,0.600,1.250,0.800,1.666,0.600,0.833,1.000,0.857,1.000,0.777,1.125,1.000,1.142,0.777,1.285,0.875,0.727,1.222,1.000,0.900,1.000,1.111,0.900,0.909,1.100,0.833,1.200,0.909,0.916,0.857,1.076,1.083,0.750$

Does there exist limit of the sequence $a(m)=\dfrac {ds_{10}(3^m)}{ds_{10}(3^{m+1})}$?

I cannot resist to note some kind of *chebyshevness* of this question (if there is one) because we know that Chebyshev proved that if limit in the prime number theorem exists then it must be equal to $1$. It could be that this is also the case here.

I also welcome any computational effort and results obtained from such an experimental work if the proof is out of reach.