91Ó°ÊÓ

If the terms' absolute values gradually decline to zero, or if $\mathbf{\left|}{\mathit{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{≤}\mathbf{\left|}{\mathit{a}}_{\mathbf{n}}\right|$and$\underset{\mathbf{n}\mathbf{→}\mathbf{∞}}{\mathbf{lim}}{\mathit{a}}_{\mathbf{n}}\mathbf{=}\mathbf{0}$, then an alternating series converges.

We may combine the positive and negative terms in this series, which is the total of two other series:

$$\begin{gathered} S=\left(2+\frac{2}{3}+\frac{2}{5}+\frac{2}{7}+...\right)+\left(-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...\right) \\ =\underset{n→1}{lim}\frac{2}{2n-1}+\underset{n→1}{lim}\frac{-1}{2n} \\ =\underset{n→∞}{lim}\frac{2}{2n-1}-\frac{1}{2n} \\ =\underset{n→∞}{lim}\frac{2\left(2n\right)-\left(2n-1\right)}{2n\left(2n-1\right)} \\ =\underset{n→∞}{lim}\frac{2n+1}{2n\left(2n-1\right)} \\ {a}_{n=}\frac{2n+1}{2n\left(2n-1\right)} \end{gathered}$$

(a) Considering step 1, See that the series is not alternating $\underset{n→∞}{lim}=\underset{n→∞}{lim}\frac{2n+1}{2n\left(2n-1\right)}$

The rule of L'Hôpital can be summarised as follows: $\underset{n→∞}{lim}=\underset{n→∞}{lim}\frac{2}{8n-2}=\frac{2}{∞}=0$

role="math" localid="1658733399079" $\underset{n→∞}{lim}{a}_n=0$

(b) Considering step 1, See that the series is not alternating$\underset{n→∞}{lim}{a}_n=\frac{n-\sqrt{n}}{n\sqrt{n}}$

Both the numerator and the denominator are divided by$\underset{n→∞}{lim}{a}_n=\underset{n→∞}{lim}\frac{1-{\displaystyle \frac{1}{\sqrt{n}}}}{\sqrt{n}}=\frac{1-{\displaystyle \frac{1}{\sqrt{∞}}}}{\sqrt{∞}}=0$

Despite satisfying role="math" localid="1658733570726" $\underset{n→∞}{lim}{a}_n=0$, the alternating series test, the series do not converge because they are not alternating series

Despite satisfying $\underset{n→∞}{lim}{a}_n=0$the alternating series test, the series do not converge since they are not alternating series (as they can be stated as the sum of two other series).