you would be surprised if i told you this

khansaheb said:
Self-doubt is the first step to salvation
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Think about it Sir khan. What logarithmic functions can give you \(\displaystyle \ln 2\)?

There are many, some are:

\(\displaystyle \ln x\) when \(\displaystyle x = 2\)
\(\displaystyle \ln(x + 1)\) when \(\displaystyle x = 1\)
\(\displaystyle \ln(x + 2)\) when \(\displaystyle x = 0\)
\(\displaystyle \ln(x + 3)\) when \(\displaystyle x = -1\)
\(\displaystyle \ln(x + 4)\) when \(\displaystyle x = -2\)
And so on...

Therefore, the most basic log function that produces \(\displaystyle \ln 2\) is \(\displaystyle \ln x\). Unfortunately, this function is not friendly to construct a Taylor series around \(\displaystyle x = 0\). So we go to the next basic one which is \(\displaystyle \ln(x + 1)\).

Let us try to write the Taylor Series Expansion.

\(\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x - c)^n\)

If \(\displaystyle c = 0\), we are constructing a Taylor Series around zero and the Taylor Series is called in this case the Maclaurin Series.

The Maclaurin Series Expansion is:

\(\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n\)

Let us do it for \(\displaystyle f(x) = \ln(x + 1)\).

\(\displaystyle f(x) = \ln(x + 1)\)
\(\displaystyle f(0) = \ln(0 + 1) = 0\)

\(\displaystyle f'(x) = \frac{1}{(x + 1)}\)
\(\displaystyle f'(0) = \frac{1}{(0 + 1)} = 1\)

\(\displaystyle f''(x) = -\frac{1}{(x + 1)^2}\)
\(\displaystyle f''(0) = -\frac{1}{(0 + 1)^2} = -1\)

\(\displaystyle f'''(x) = \frac{2}{(x + 1)^3}\)
\(\displaystyle f'''(0) = \frac{2}{(0 + 1)^3} = 2\)

\(\displaystyle f^{4}(x) = -\frac{6}{(x + 1)^4}\)
\(\displaystyle f^{4}(0) = -\frac{6}{(0 + 1)^4} = -6\)

\(\displaystyle f^{5}(x) = \frac{24}{(x + 1)^5}\)
\(\displaystyle f^{5}(0) = \frac{24}{(0 + 1)^5} = 24\)

Then,

\(\displaystyle \ln(x + 1) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = \frac{0}{0!}x^0 + \frac{1}{1!}x^1 - \frac{1}{2!}x^2 + \frac{2}{3!}x^3 - \frac{6}{4!}x^4 + \frac{24}{5!}x^5 - \cdots\)

\(\displaystyle \ln(x + 1) = x - \frac{1}{2}x^2 + \frac{2}{6}x^3 - \frac{6}{24}x^4 + \frac{24}{120}x^5 - \cdots\)

\(\displaystyle \ln(x + 1) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 - \cdots\)

Do you see a pattern?

\(\displaystyle \ln(x + 1) = \sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}x^n\)

When \(\displaystyle x = 1\), we get:

\(\displaystyle \ln(1 + 1) = \sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}1^n = \sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n} = \ln 2\)

So, there is no magic why this series converges to \(\displaystyle \ln 2\).

💪😏😏
 
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The "surprise" in this context is Riemann's series theorem: You can always rearrange the terms so that it converges to any given real number.
 
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