By the test of reason, this series converges. The problem is figuring out which technique to use to calculate your sum.
Thanks for any help.
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3I doubt this series has a closed form. – Connor Harris Feb 12 '19 at 21:30
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Good question. I do not know if there is a way. I had not thought of that. – Mathsource Feb 12 '19 at 21:32
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9The test of reason is my favorite convergence test. – Antonio Vargas Feb 12 '19 at 21:33
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6@AntonioVargas, I'm guessing that the OP means "ratio test." The word "rational" has two very different meanings and this probably led to a poor translation. – Cheerful Parsnip Feb 12 '19 at 21:35
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1Wolfram alpha tells that the limit of the sum is $≈1,27498$: https://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B1%7D%7B3%5En-2%5En%7D%7D – Dr. Mathva Feb 12 '19 at 22:23
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You can express the sum in terms of the $q$-digamma function $\psi_q(z)$. This isn't a profound simplification, however, because $\psi_q(z)$ is defined as a sum of a similar form: $$ \psi_q(z) = \frac{\partial \log \Gamma_q(z)}{\partial z} = -\log(1-q) + \log(q) \sum_{n=0}^\infty \frac{q^{n+z}}{1-q^{n+z}}. $$ But, for what it's worth, $$ \sum_{n=1}^\infty \frac{1}{3^n-2^n} = \frac{\log3-\psi_{2/3}(\log_{3/2}3)}{\log(3/2)}. $$
Jim Ferry
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$\Gamma_q(z)$ is the $q$-gamma function. There are many functions that have "$q$-analogs," and I included the reference to $\Gamma_q(z)$ in order to emphasize that the definition of $\psi_q(z)$ isn't some strange, arbitrary thing, but rather the derivative of $\log \Gamma_q(z)$, in analogy to the definition of the standard digamma function. – Jim Ferry Feb 12 '19 at 22:48