- #1
murshid_islam
- 457
- 19
- Homework Statement
- To determine whether a series is convergent using the Alternating Series Test
- Relevant Equations
- [tex]\sum\limits_{n=1}^{\infty} (-1)^n\frac{\sqrt n}{2n+3}[/tex]
The Alternating series test has to be used to determine whether this series converges or diverges: [tex]\sum\limits_{n=1}^{\infty} (-1)^n\frac{\sqrt n}{2n+3}[/tex]
Here's what I have done:
Let [itex]a_n = \frac{\sqrt n}{2n+3}[/itex]. Therefore, [itex]a_{n+1} = \frac{\sqrt {n+1}}{2n+5}[/itex]
Now, for [itex]a_{n+1}[/itex] to be less than or equal to [itex]a_n \\ [/itex],
[itex]\frac{\sqrt {n+1}}{2n+5} \leq \frac{\sqrt n}{2n+3} \\ [/itex]
[itex]\Rightarrow \sqrt {\frac{n+1}{n}} \leq \frac{2n+5}{2n+3} < \frac{4n+6}{2n+3} \\ [/itex]
[itex]\Rightarrow \sqrt {\frac{n+1}{n}} < 2 \\ [/itex]
[itex]\Rightarrow \frac{n+1}{n} < 4 \\ [/itex]
[itex]\Rightarrow 3n > 1 \\ [/itex]
[itex]\Rightarrow n > \frac{1}{3}[/itex]
Therefore, [itex]a_{n+1} \leq a_n[/itex] for all [itex]n \geq 1[/itex]
And [itex]\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\sqrt n}{2n+3} = \lim_{n \to \infty} \frac{1}{2 \sqrt n+ \frac{3}{\sqrt n}} = 0[/itex]
Therefore, by Alternating Series Test, the series is convergent.
Is this ok?
.
Here's what I have done:
Let [itex]a_n = \frac{\sqrt n}{2n+3}[/itex]. Therefore, [itex]a_{n+1} = \frac{\sqrt {n+1}}{2n+5}[/itex]
Now, for [itex]a_{n+1}[/itex] to be less than or equal to [itex]a_n \\ [/itex],
[itex]\frac{\sqrt {n+1}}{2n+5} \leq \frac{\sqrt n}{2n+3} \\ [/itex]
[itex]\Rightarrow \sqrt {\frac{n+1}{n}} \leq \frac{2n+5}{2n+3} < \frac{4n+6}{2n+3} \\ [/itex]
[itex]\Rightarrow \sqrt {\frac{n+1}{n}} < 2 \\ [/itex]
[itex]\Rightarrow \frac{n+1}{n} < 4 \\ [/itex]
[itex]\Rightarrow 3n > 1 \\ [/itex]
[itex]\Rightarrow n > \frac{1}{3}[/itex]
Therefore, [itex]a_{n+1} \leq a_n[/itex] for all [itex]n \geq 1[/itex]
And [itex]\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\sqrt n}{2n+3} = \lim_{n \to \infty} \frac{1}{2 \sqrt n+ \frac{3}{\sqrt n}} = 0[/itex]
Therefore, by Alternating Series Test, the series is convergent.
Is this ok?
.