Special Series — Question 5

Question 5 – Solution

Step 1: Prove convergence.

Let $b_n=1/n$. Then $b_n>0$, $b_{n+1}<b_n$, and $b_n\to0$. Therefore the Alternating Series Test proves that $\sum(-1)^{n-1}b_n$ converges.

Step 2: Test absolute convergence.

Taking absolute values gives $$\sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}n\right| =\sum_{n=1}^{\infty}\frac1n,$$ the divergent harmonic series. Hence the original series is conditionally convergent, not absolutely convergent.

Step 3: Identify the sum.

For $-1<x\le1$, the endpoint-valid power-series identity is $$\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}.$$ Setting $x=1$ gives $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n=\ln2.$$

Step 4: Quantify the approximation.

The Alternating Series Estimation Theorem gives $|S-s_N|\le b_{N+1}=1/(N+1)$. Odd partial sums lie above $\ln2$ and even partial sums lie below it.