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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.

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there is no $x$ in the expression that you have there as is. – deoxygerbe Apr 28 '12 at 12:35
up vote 1 down vote accepted

If your series is actually $$\sum_{n\ge 0}2^n\ln\left(1+\frac{x}{3^n}\right)\;,$$ your ratio should be $$\frac{2\ln\left(1+\frac{x}{3^{n+1}}\right)}{\ln\left(1+\frac{x}{3^n}\right)}\;.$$ You can use l’Hospital’s rule to find the limit of this as $n\to\infty$:

$$\begin{align*} \lim_{n\to\infty}\frac{2\ln\left(1+\frac{x}{3^{n+1}}\right)}{\ln\left(1+\frac{x}{3^n}\right)}&=\lim_{n\to\infty}\frac{2\left(1+\frac{x}{3^{n+1}}\right)^{-1}(-x\ln 3)3^{-(n+1)}}{\left(1+\frac{x}{3^n}\right)^{-1}(-x\ln 3)3^{-n}}\\\\ &=\frac23\lim_{n\to\infty}\frac{1+\frac{x}{3^n}}{1+\frac{x}{3^{n+1}}}\\\\ &=\frac23\;. \end{align*}$$

(I didn’t bother with the absolute values, since everything here is positive anyway.)

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De l'Hospital is not for sequences. I mean that your solution is not correct but may be easily corrected :-) – Siminore Apr 28 '12 at 13:07
@Siminore Heine-Borel Theorem.I think he means that – 89085731 Apr 28 '12 at 13:10
@Siminore: On the contrary, it is correct. It’s a standard result that l’Hospital’s rule can be applied to sequences when the terms are defined by a function to which it applies. – Brian M. Scott Apr 28 '12 at 13:11
@Gingerjin: I meant exactly what I wrote. – Brian M. Scott Apr 28 '12 at 13:11
@Siminore: I would say in fact that it’s an obvious difference, not a subtle one. But I repeat: this application of l’Hospital’s rule and its justification are standard fare in first-year calculus courses, so there is nothing unfair about using it without comment. – Brian M. Scott Apr 28 '12 at 14:08

Assuming, as I deduce from you post, that the series is $$\sum_n 2^n \log \left(1+\frac{x}{3^n} \right),$$ for each fixed $x$ it is asymptotic to $$\sum_n 2^n \frac{x}{3^n},$$ which is always convergent. I have used the fact that $\log(1+\varepsilon) \sim \varepsilon$ as $\varepsilon \to 0$.

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