Have the following:
$\sum 2^{n}\log(1+\frac{1}{3^{n}})$
Now I was thinking the best way to approach would be via the ratio test, doing so I got to the following,
$\rvert\frac{a_{n+1}}{a_n}\rvert= \rvert \log(1+\frac{x}{3^{n+1}})\rvert$ Hence then using the fact that for this to converge it must be less then one and given the x>0, we have that x<$3^{n+1}(e-1)$. Not sure if i'm going down the correct route here, any help would be much appreciated.