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$ 1 + \frac 13 . \frac 14 + \frac 15 . \frac 1{4^2} + \frac 17 . \frac 1{4^3} + ...$

I evaluated the expression for the first few terms and I find that this number will probably tend to $ \log 3 $. I'd like to know why, or how I can prove that it does indeed tend to $ \log 3 $. More importantly, I'd like to know the relationship between this particular type of a series and the natural logarithm of numbers.

Why does the natural logarithm of a number show up in such a series?

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marked as duplicate by Martin R, Did, José Carlos Santos sequences-and-series Jan 3 at 17:59

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  • $\begingroup$ Looks like: $$\sum_{r=0}^{\infty}{\frac{1}{4^r(2r+1)}}$$ $\endgroup$ – Rhys Hughes Jan 3 at 17:59
  • $\begingroup$ The general term is $[(2n+1)4^n]^{-1}$ if that helps. $\endgroup$ – John Jan 3 at 17:59
  • $\begingroup$ The reason $\log$ appears is simple: its derivative is $\frac1x$, which is exactly what allows you to get all the odd numbers to appear in the denominators.. $\endgroup$ – Arthur Jan 3 at 18:00
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    $\begingroup$ You should be careful about writing $+\dots\infty$, it looks like you're adding infinity to all the other fractions. This, however, is not what you mean. $\endgroup$ – Michael Burr Jan 3 at 18:02
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Your series is: $$f(x) = \sum_{n=0}^\infty\dfrac{x^n}{2n+1}$$ with $x = \dfrac{1}{4}.$ The standard way to go about this kind of problem is to observe that $f(x)$ only converges uniformly for $|x|\leq r<1$ for any such $r$ and then manipulate the series into something we know already, usually involving taking derivatives.

If you want the challenge, then let me give you a hint: For $0<x<1$, consider $xf(x^2)$ and then take its derivative to get something you can easily calculate.

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  • $\begingroup$ @J.G. you are right, that was a mistake. $\endgroup$ – dezdichado Jan 3 at 18:01

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