<s>[INST] What is the sum of series $1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$ I am trying to solve this series question which is 
$$1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$$
Till now I've only been able to write the general form of this series which is 
$$\sum_{n=1}^\infty\frac1{(2n-1)4^{n-1}}$$
but I don't know how to proceed. 
 [/INST] Hint Reindexing your expression for the series for convenience gives
$$\sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\frac{1}{4}\right)^k .$$
Now, the factors $\frac{1}{2 k + 1}$ suggest considering the power series
$$\operatorname{artanh} x \sim \sum_{k = 0}^\infty \frac{1}{2 k + 1} x^{2 k + 1} .$$

Rewrite the given series to get $$\sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\frac{1}{4}\right)^k = 2 \sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\color{red}{\frac{1}{2}}\right)^{2 k + 1} = 2 \operatorname{artanh} \color{red}{\frac{1}{2}} = 2 \cdot \frac{1}{2}\log \frac{1 + \color{red}{\frac{1}{2}}}{1 - \color{red}{\frac{1}{2}}} = \boxed{\log 3} .$$

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