Series not in the Video

[nasi112]

[math]\sum_{n=0}^{\infty}(-1)^{n+1}\frac{\sqrt{n}}{n + 1}[/math]
If I use the alternating series test, I need to show [imath]\frac{\sqrt{n + 1}}{n + 2} \leq \frac{\sqrt{n}}{n + 1}[/imath] for all [imath]n \geq 0[/imath].

[math]\frac{\sqrt{n + 1}}{n + 2}\frac{n + 1}{\sqrt{n}} \leq 1[/math]
[math]\frac{(n + 1)^{3/2}}{n^{3/2} + 2n^{1/2}} \leq 1[/math]
It's not clear to me this inequality is true for all \(\displaystyle n \geq 1\).

 

[lookagain]

If I use the alternating series test, I need to show [imath]\frac{\sqrt{n + 1}}{n + 2} \leq \frac{\sqrt{n}}{n + 1}[/imath] for all [imath]n \geq 0[/imath].

It is not true when n equals 0. Start with n equals 1.

It's the first fraction versus the second fraction. Square each side:

\(\displaystyle \dfrac{n + 1}{(n + 2)^2} \ \ \ vs. \ \ \ \dfrac{n}{(n + 1)^2} \)

\(\displaystyle (n + 1)^2(n + 1) \ \ \ vs. \ \ \ (n + 2)^2(n) \)

\(\displaystyle (n + 1)^3 \ \ \ vs. \ \ \ (n + 2)^2(n)\)

\(\displaystyle n^3 + 3n^2 + 3n + 1 \ \ \ vs. \ \ \ n^3 + 4n^2 + 4n \)

\(\displaystyle 1 \ \ \ vs. \ \ \ n^2 + n \)

. . .

 

[nasi112]

Thanks for noticing this. I'll rewrite it.

If I use the alternating series test, I need to show [imath]\frac{\sqrt{n + 1}}{n + 2} \leq \frac{\sqrt{n}}{n + 1}[/imath] for all [imath]n \geq 1[/imath].

 

[nasi112]

Thanks. Got it. Awesome comparison.