Find the sum sum from n=1 to infinity frac (-1)^(n-1) n

Find the sum $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n$.

Explanation of Method

This is exactly the $\ln(1+x)$ Maclaurin series evaluated at $x=1$; since that series converges (conditionally) at the endpoint and $\ln(1+x)$ is continuous there, plugging in $x=1$ gives the sum directly.

Recognize the series as $\ln(1+x)$ at $x=1$.
Confirm convergence at the endpoint via the Leibniz test.
Verify numerically with partial sums.

Step 1 · Identify the Taylor series

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

Step 2 · Confirm convergence

With $a_n=1/n$: positive, decreasing, $\to0$ — the Leibniz alternating series test applies, and continuity of $\ln(1+x)$ at $x=1$ guarantees the series converges to $\ln(1+1)=\ln2$, not merely that it converges to some value.

Step 3 · Numerical check

$$ S_1=1,\ \ S_2=0.5,\ \ S_3=0.8333,\ \ S_{10}\approx0.6456,\ \ S_{100}\approx0.6882,\ \ S_{1000}\approx0.6926, $$ slowly approaching $\ln2\approx0.693147$.

Summary Table

Property	Value
Sum	$\ln2\approx0.693147$
Convergence type	Conditional
Origin	$\ln(1+x)$ at $x=1$

Additional Notes

This series converges only conditionally, not absolutely (the absolute-value series is the divergent harmonic series) — a classic example distinguishing the two notions.
Because convergence is conditional, the Riemann rearrangement theorem applies: reordering the terms can make the series converge to any real number, or diverge, unlike an absolutely convergent series.
The slow convergence (needing roughly $10^5$ terms for 5-decimal accuracy) is exactly why Problem 279's error-bound analysis and Machin-type acceleration tricks (Problem 282) matter in practice.

Result

$$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n = \ln2. $$