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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.2012-04-28

2 Answers 2

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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 :-)2012-04-28
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    @Siminore Heine-Borel Theorem.I think he means that2012-04-28
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    @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.2012-04-28
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    @Gingerjin: I meant exactly what I wrote.2012-04-28
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    Yes sorry it should have been an x in there and asks for which values does the series converge, so am i correct in thinking it converges to 2/3 or is this the ratio?2012-04-28
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    @user24930: $2/3$ is the limit of the ratio of consecutive terms. Since $|2/3|<1$, the series converges for all $x>0$.2012-04-28
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    @BrianM.Scott many thanks for that.2012-04-28
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    @BrianM.Scott: there is a subtle difference between $f \colon \mathbb{R} \to \mathbb{R}$ and $f_{|\mathbb{N}} \colon \mathbb{N} \to \mathbb{R}$. The second one cannot be differentiated, so that it is a bit unfair to write something like $$\lim_{n \to +\infty} \frac{f(n)}{g(n)} = \lim_{n \to +\infty} \frac{f'(n)}{g'(n)}.$$ You'd better claim that you compute the limit with a continuous variable and then remark that the limit with the discrete variable must coincide. But I agree this is rather pedantic.2012-04-28
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    @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.2012-04-28
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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 $\varepsilon \to 0$.