# Summation of logarithmic functions

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?

• Maybe a Taylor series? – Andrew Li Apr 27 '18 at 3:22

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 $$\sinh \log 3 = \frac{e^{\log 3}-e^{-\log 3}}{2} = \frac{3-1/3}{2} = \boxed{\frac{4}{3}}\,.$$
• @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. – Clement C. Apr 27 '18 at 22:08
• 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 $f=\sinh$. – Clement C. Apr 27 '18 at 22:16