Taylor series expansion of $\log{x}$ to obtain infinite sequence for $\log{2}$

So I know that $\log{2} = 1 - 1/2 + 1/3 - 1/4 + 1/5 ...$

I tried to arrive at this result by using Taylor Series Expansion on $\log{x}$ around $1$, my thinking was that if infinitely many terms are used in the expansion then once I substitute in for $2$, I will get the mentioned series. However, I actually obtained:

$$\log{2} = 1 - 1/2! + 1/3! - 1/4! + 1/5! ...$$

And I noticed that to arrive at the correct result, one has to perform Taylor Series Expansion on $\log{1-x}$ around $0$. Which also makes sense. My question is: why performing TSE on $\log{x}$ around $1$ does not yield the correct result?

EDIT: My TSE

$$f(x) = \frac{f^{(i)}(x_0)}{i!}(x-x_0)^i$$

$$f^{(0)}(x) = \log{x}$$

$$f^{(1)}(x) = x^{-1}$$ $$f^{(2)}(x) = -x^{-2}$$ $$f^{(3)}(x) = 2x^{3}$$

and so TSE around $1$

$$f(x) = \log{1} + \frac{1^{-1}}{1!}(x-1) - \frac{1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 + ...$$

$$f(x) = 0 + x-1 - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3}$$

$$f(2) = 1 - \frac{1}{2} + \frac{1}{3} ...$$

Oh dear... Looks like I messed up my TSE before

• Isn't $\log(1+x)=x-x^2/2+x^3/3-x^4/4+\cdots$ without factorials? Sep 28, 2016 at 19:29
• yes, it is without factorials. But when I perform TSE on $\log{x}$ around $1$, I get factorials... Am I messing something up.. Should I write out my TSE?
– Naz
Sep 28, 2016 at 19:30
• Let $f = \log$. Then for $k \geqslant 1$ we have $$f^{(k)}(x) = \frac{(-1)^{k-1}(k-1)!}{x^k}.$$ Seems you missed the $(k-1)!$. Sep 28, 2016 at 19:31
• @DanielFischer Yes, you are right
– Naz
Sep 28, 2016 at 19:38
• $f^{(n)}(x)=(-1)^{n-1}(n-1)!x^{-n}$ with $f(x)=ln(x)$ and n>0. Sep 28, 2016 at 19:51

Instead of considering $$\tag{1}\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$$ or $$\tag{2}\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-...$$

it is classical, for obtaining a better convergence, to consider their difference (1)-(2) giving

$$\tag{3}\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$

Setting $$x=\frac{1}{3}$$ in (3) gives the series:

$$\tag{4}\ln(2)=2\left(\frac{1}{3}+\frac{1}{(3 \times 3^3)}+\frac{1}{(5 \times 3^5)}+\frac{1}{(7 \times 3^7)}+...\right)$$

which is quite rapidly converging: with only the four terms given in (4), we get $$\ln(2)\approx 0.693135...$$, the true value being $$0.693147...$$, a $$10^{-5}$$ error approximately.

Remark: I just discovered (2022/03/28) a Wikipedia entry specificaly devoted to $$\ln(2)$$ listing many astound series...

Because there are many expansions for the logarithm, with respect to the point you're expanding at, or better: the range.

Here are the most famous log expansion, with their ranges:

$$\ln(1+x) = \sum_{k = 1}^{+\infty} (-1)^{k+1}\frac{x^k}{k} ~~~~~~~ \text{for}\ -1 < x \leq +1$$

$$\ln(x) = \sum_{k = 1}^{+\infty} (-1)^{k+1}\frac{(x-1)^k}{k} ~~~~~~~ \text{for}\ 0 < x \leq +2$$

$$\ln(x) = 2\sum_{k = 1}^{+\infty} \frac{1}{2k-1}\left(\frac{x-1}{x+1}\right)^{2k-1} ~~~~~~~ \text{for}\ 0 < x$$

$$\ln(x) = \sum_{k = 1}^{+\infty} \frac{1}{k}\left(\frac{x-1}{x}\right)^k ~~~~ \text{for}\ x \geq +\frac{1}{2}$$

In any case, you missed a factorial!

• how are those ranges determined? i.e. where do they come from?
– Naz
Sep 28, 2016 at 19:48
• @isquared-KeepitReal You can find a complete treatise of those series in almost every Calculus book. Basically it's about determining the radius of convergence of each series! Sep 28, 2016 at 19:57