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I know what the Taylor Series of $\ln(1+x)$ is, but, I don't know why it's true also for $x=1$.

If I decide to use the method of integration starting from $\frac{1}{1+x} \ =\ \sum ( -1)^{n} x^{n}$

then, that is true only for $-1< x< 1$.

How can I prove it's also for $x=1$?

Thank you!

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    $\begingroup$ It is not true for $x=1$ because the taylor series for $\ln(1+x)$ is defined for $|x|<1$ $\endgroup$
    – Eminem
    May 30, 2020 at 17:09
  • $\begingroup$ That is not what is written here: math.stackexchange.com/questions/878374/taylor-series-of-ln1x last comment on the most voted answer. Plus, I was asked to prove it, so I assume it must be true. $\endgroup$
    – saraspagno
    May 30, 2020 at 17:16
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    $\begingroup$ It is true for x=1, it's the alternating harmonic series. In this specific case you can just test whether it converges at x=1 and x=1 (the alternating series test will show that it does indeed converge at x=1) but it can be shown to not converge for x=-1 (normal harmonic series). Not sure about more general results for arbitrary series. Have a look into 'radius of convergence' $\endgroup$
    – Joz
    May 30, 2020 at 17:17
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    $\begingroup$ Thank you! I know if I plug in $x=1$ the series converges. But how does it show the equality that $ln(2) = ln(1+x)$. How is this connected to showing that it's Taylor series converges? Shouldn't we first show that the equality holds when $x=1$? $\endgroup$
    – saraspagno
    May 30, 2020 at 17:19
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    $\begingroup$ Look up Abel's theorem $\endgroup$
    – Gary
    May 30, 2020 at 17:20

1 Answer 1

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Since the series $\frac{1}{1+x} \ =\ \sum ( -1)^{n} x^{n}$ is valid for $|x|\lt 1$ and you have obtained Taylor series by integrating the series, it may seem that the Taylor series for $\ln (1+x)$ is not valid at $x=1$. One way to be sure that the Taylor series is indeed valid at $x=1$, is to go by definition.

Let $f(x)=\frac{1}{1+x}$. We want to write Taylor series for $f$ about $x=0$ in the domain $[0,x]$ where $x\gt 0$. Find higher order derivatives for $f$, you'll notice that $f^{(n)}(x)=(-1)^n \frac{n!}{(1+x)^n}$

$f(x)=\ln (1+x)= f(0)+\sum_{i=1}^{i=n-1}\frac{f^{(i)}(0)}{i!}x^i + \frac{f^{(n)}(c)}{n!}x^n$, where $c\in (0,x)$.
$\implies f(x)=\sum_{i=1}^{i=n-1} (-1)^ix^i + \frac{(-1)^n}{(1+c)^n}x^n=P_{n-1}(x)+R_{n-1}(x)$, where $R_{n-1}(x)=\frac{(-1)^n}{(1+c)^n}x^n$ is remainder term.

Note that, $|R_{n-1}(x)|=\frac{1}{(1+c)^n}x^n \to 0$ as $n\to \infty$ for $ x=1$
Hence, $P_{n-1}$ converges to $f(x)$ when $x=1$

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