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
$\endgroup$ 52 Answers
$\begingroup$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...
$\endgroup$ $\begingroup$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!
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