---
tag: VAV1B60704
kind: proof
name: Proof of the Mercator series by term-by-term differentiation and Abel's theorem
msc: [30B10, 33B10]
proves: 0YZXNE3HFR
uses: [0044CPHJBV, 07X1QA6BRW, 0RXX2PT1MQ, 0MH1WK89PP, 0JBMYAA9QH, 002X7EK91H, 01X5CVV1Q5, 003QDZ4T16, 0RKW2X63YR, 0PK8QH1W1V, 08QAX4H1PF, 0S4EMDQS2F, 0ZXJ2455NJ, 0RMVS7RXZX]
sources:
  - url: https://en.wikipedia.org/wiki/Mercator_series
    locator: Derivation
  - url: https://en.wikipedia.org/wiki/Abel%27s_theorem
---

## Proof

We first treat $-1 < x < 1$ by comparing derivatives, then the endpoint $x = 1$ by Abel's theorem.

1. Let $a_0 = 0$ and $a_n = \frac{(-1)^{n+1}}{n}$ for $n \ge 1$. Consider the power series $g(x) = \sum_{n=0}^{\infty} a_n x^n$.
2. For $n \ge 1$, $\frac{|a_{n+1}|}{|a_n|} = \frac{n}{n+1} \to 1$. By the [[ratio formula|0JBMYAA9QH]], the radius of convergence of $g$ is $1$.
3. By [[term-by-term differentiation|002X7EK91H]], $g$ is differentiable on $(-1, 1)$. Its derivative is $g'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1}$.
4. Putting $m = n - 1$, this is $g'(x) = \sum_{m=0}^{\infty} (-x)^m$. Since $\abs{-x} < 1$, the [[geometric series|01X5CVV1Q5]] gives $g'(x) = \frac{1}{1 + x}$.
5. Let $h(x) = \ln(1 + x)$ for $-1 < x < 1$, where $\ln$ is the [[natural logarithm|0044CPHJBV]]. This is defined, since $1 + x > 0$.
6. The map $x \mapsto 1 + x$ is a [[polynomial with derivative $1$|0MH1WK89PP]], and it sends $(-1, 1)$ into $(0, \infty)$.
7. By the [[derivative of the logarithm|07X1QA6BRW]] and the [[chain rule|0RXX2PT1MQ]], $h$ is differentiable on $(-1, 1)$ with $h'(x) = \frac{1}{1 + x} \cdot 1$.
8. So $(g - h)'(x) = 0$ for every $x$ in the open interval $(-1, 1)$. By [[this result|003QDZ4T16]], $g - h$ is constant on $(-1, 1)$.
9. At $x = 0$, $g(0) = a_0 = 0$ and $h(0) = \ln 1 = 0$, since $e^0 = 1$ by the [[series of the exponential|0RMVS7RXZX]] and [[$\ln(e^y) = y$|0RKW2X63YR]] for real $y$. So $g = h$ on $(-1, 1)$.
10. This proves $\ln(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n$ for $-1 < x < 1$.
11. Now let $x = 1$. Let $b_k = \frac{1}{k+1}$ for $k \ge 0$. Then $(b_k)$ is decreasing, and [[$b_k \to 0$|08QAX4H1PF]].
12. By the [[alternating series test|0PK8QH1W1V]], $\sum_{k=0}^{\infty} (-1)^k b_k$ converges. With $n = k + 1$, this is the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$. Call its sum $S$.
13. Then $\sum_{n=0}^{\infty} a_n = S$ converges, since $a_0 = 0$. By [[Abel's theorem|0S4EMDQS2F]], $g(x) \to S$ as $x \to 1$ from the left.
14. By step 9, $g(x) = \ln(1 + x)$ for $0 < x < 1$. Since [[$\ln$ is continuous|0RKW2X63YR]] at $2$, $\ln(1 + x) \to \ln 2$ as $x \to 1$ from the left.
15. Take any sequence $x_k \in (0, 1)$ with $x_k \to 1$. Then $g(x_k) \to S$ and $g(x_k) \to \ln 2$. By [[uniqueness of limits|0ZXJ2455NJ]], $S = \ln 2$.
16. So $\ln(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n$ for every $x$ with $-1 < x \le 1$.

## Idea

Inside the interval of convergence, the series and $\ln(1 + x)$ have the same derivative $\frac{1}{1 + x}$, by the geometric series. They also agree at $0$, so they are equal. At the endpoint $x = 1$ the series still converges, and Abel's theorem says its sum is the limit of the values inside, which is $\ln 2$.