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The sum of series $\frac{(\log3)^1}{1!}+\frac{(\log3)^3}{3!}+\frac{(\log 3)^5}{5!}+\cdots$ is what? Is there a general algorithm to find the summation of logarithms?

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    $\begingroup$ Maybe a Taylor series? $\endgroup$ – Andrew Li Apr 27 '18 at 3:22
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The series itself does not have too much to do with logarithms; to see why without getting lost with the $\log$ everywhere, let $\alpha = \log 3$. You want $$ \sum_{n=0}^\infty \frac{(\log 3)^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty \frac{\alpha^{2n+1}}{(2n+1)!} = \sinh \alpha $$ by the series definition of $\sinh$. That being said, now here we have simplifications because $\alpha=\log 3$. Indeed, recall that, for every $x\in\mathbb{R}$, $$ \sinh x = \frac{e^x-e^{-x}}{2} $$ and therefore here $$ \sinh \log 3 = \frac{e^{\log 3}-e^{-\log 3}}{2} = \frac{3-1/3}{2} = \boxed{\frac{4}{3}}\,. $$

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    $\begingroup$ I was wondering what taylor series it was, forgot about hyperbolics. $\endgroup$ – Andrew Li Apr 27 '18 at 3:30
  • $\begingroup$ @ClementC: if we don't recognize which Taylor Series, what is a good procedure? This one is odd, and tends to +/-infinity. $\endgroup$ – smci Apr 27 '18 at 22:06
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    $\begingroup$ @smci You always have the option of trying a formal math software, e.g. Mathematica or Wolfram Alpha, to get a hint "recognizing" the series. But otherwise, here, would you have recognized it with a $(-1)^n$ in each term? (to get $\sin$) If so, then recognizing $\sinh$ is not too hard. Or you could directly recognize something like "only the odd indices of $\exp$", so that thinking of $\frac{e^{x}-e^{-x}}{2}$ (the odd part of $\exp$) is natural. $\endgroup$ – Clement C. Apr 27 '18 at 22:08
  • $\begingroup$ Or if you really don't recognize anything, try to find some differential equation the series satisfies (differentiating to get termwise derivatives, then reorganizing); solving the differential equation would in many case allow you to find the corresponding function. $\endgroup$ – Clement C. Apr 27 '18 at 22:09
  • $\begingroup$ For instance, here differentiating twice the power series we get $$f''(x) = \sum_{n=0}^\infty (2n+1)2n \frac{x^{2n-1}}{(2n+1)!} \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!} = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!} = f(x)$$ so that solving $f''=f$ with $f(0)=0$ and $f'(0)=1$ will give you the solution $f=\sinh$. $\endgroup$ – Clement C. Apr 27 '18 at 22:16

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