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I am currently struggling with finding the power series, convergence radius, and interval for the following functions:

a) $f(x)=\frac{2}{1-x}$

b) $f(x)=2 \ln (1-x)$

Here is what I have attempted so far:

a) For function a), I started by using the formula for the geometric series to write:

$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$

Then, I multiplied both sides by 2 to get:

$$\frac{2}{1-x} = \sum_{n=0}^{\infty} 2x^n$$

This gives me the power series for $f(x)$, but I am unsure how to find the convergence radius and interval.

b) For function b), I used the power series expansion for $\ln (1-x)$, which is:

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

Then, I multiplied both sides by 2 to get:

$$2 \ln (1-x) = -2 \sum_{n=1}^{\infty} \frac{x^n}{n}$$

Again, I am not sure how to find the convergence radius and interval for this function.

Could someone please help me with finding the convergence radius and interval for these functions? Any explanation or guidance would be greatly appreciated.

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    $\begingroup$ The radius of convergence $R$ of these two power series is the same (since one is, up to sign, the derivative of the other one), and for the first one (geometric series) it is well-known: $R=1.$ The interval is $(-R,R).$ More precisely, both are abolutely convergent on this open interval. The geometric series is trivially divergent for $x=\pm R.$ $\endgroup$
    – Anne Bauval
    Commented Feb 25, 2023 at 17:26
  • $\begingroup$ @Anne Bauval I see, so since the power series for the two functions are related to each other by a constant multiple, they will have the same radius of convergence. That makes sense. I also appreciate the reminder that the radius of convergence for a geometric series is 1. Therefore, for function a), the radius of convergence is 1 and the interval of convergence is (-1,1). And since the power series for function a) converges absolutely for all $x$ in the interval (-1,1), the interval of convergence is also the interval of absolute convergence. $\endgroup$
    – Bryan C
    Commented Feb 25, 2023 at 17:31
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    $\begingroup$ Yes, and a finer result (not deducible from the previous ones) is that the series for $\ln(1-x)$ is divergent for $x=1$ but semi-convergent (alternate) for $x=-1.$ $\endgroup$
    – Anne Bauval
    Commented Feb 25, 2023 at 17:33
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    $\begingroup$ "The series for $\ln(1-x)$ is divergent when $x=1$" not because "the terms of the series do not approach zero as n approaches infinity" (they do!) but because it is (up to sign) the harmonic series. Note also that the last equation in your comment is not obvious. You can use Taylor-Lagrange to prove it. $\endgroup$
    – Anne Bauval
    Commented Feb 25, 2023 at 17:39
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    $\begingroup$ One way to prove it rigorously is by using Taylor's theorem with the Lagrange remainder. Specifically, we can write: $$\ln(1-x) = -\sum_{n=1}^{N-1} \frac{x^n}{n} - \frac{x^N}{N} \cdot \frac{1}{1-t} \cdot \frac{(-x)^N}{N}$$ for some $t$ between $x$ and 0, since the Lagrange remainder term for the Taylor series expansion of ln(1-x) is given by: $$R_N(x) = \frac{(-x)^N}{N(1-t)^{N+1}}$$ Taking the limit as $N \rightarrow \infty$, we have: $$\ln2 = -\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$$ which is the well-known result for the alternating harmonic series. $\endgroup$
    – Bryan C
    Commented Feb 25, 2023 at 17:47

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