Question

Determine whether the series in Exercises 21–23 converge or diverge. Note that these series do not satisfy the criteria of the alternating series test.

$$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+...$$

Short answer

The given series converges.

Step-by-step solution

Step 1. Given Information

• Given series, $$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+...$$
• These series do not satisfy the criteria of the alternating series test.

Step 2. Explanation

Here, the given series can be defined as,

$$u_{n}=(-1)^{n}(\frac{1}{2n}+\frac{1}{2n+1})$$

$$\implies u_{n}=(-1)^{n}(\frac{4n+1}{4n^{2}+2n})$$

Using the ratio test, we get

$$\lim_{n \to \infty}\frac{u_{n}+1}{u_{n}}=\lim_{n \to \infty}(\frac{(-1)^{n+1}(4n+5)}{(2n+2)(2n+3)}\frac{(2n)(2n+1)}{(-1)^{n}(4n+1)})$$

$$\implies \lim_{n \to \infty}\frac{u_{n}+1}{u_{n}}=-1$$

SInce, $$-1<1$$, the ratio test shows the series converging.

Hence, the given series converges.