# Why does $1 + \frac 13 . \frac 14 + \frac 15 . \frac 1{4^2} + \frac 17 . \frac 1{4^3} + … = \log 3$? [duplicate]

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

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?

## marked as duplicate by Martin R, Did, José Carlos Santos sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 3 at 17:59

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• Looks like: $$\sum_{r=0}^{\infty}{\frac{1}{4^r(2r+1)}}$$ – Rhys Hughes Jan 3 at 17:59
• The general term is $[(2n+1)4^n]^{-1}$ if that helps. – John Jan 3 at 17:59
• 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.. – Arthur Jan 3 at 18:00
• 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. – Michael Burr Jan 3 at 18:02

## 1 Answer

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 $$0, consider $$xf(x^2)$$ and then take its derivative to get something you can easily calculate.

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