MHB Checking Convergence of Series: Inequalities & Tests

[mathmari]

Hey!

I want to check the convergence for the below series.

- $\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$

Let $\displaystyle{a_n=\frac{\left (n!\right )^2}{\left (2n+1\right )!}\cdot 4^n}$.
Then we have that \begin{align*}a_{n+1}&=\frac{\left ((n+1)!\right )^2}{\left (2(n+1)+1\right )!}\cdot 4^{n+1}=\frac{(n!)^2(n+1)^2}{\left (2n+3\right )!}\cdot 4^{n}\cdot 4=\frac{(n!)^2(n+1)^2}{\left (2n+1\right )!(2n+2)(2n+3)}\cdot 4^{n}\cdot 4 \\ & =\frac{(n!)^2(n+1)^2}{\left (2n+1\right )!2(n+1)(2n+3)}\cdot 4^{n}\cdot 4 =\frac{(n!)^2(n+1)}{\left (2n+1\right )!(2n+3)}\cdot 4^{n}\cdot 2\end{align*}
So we get \begin{equation*}\frac{a_{n+1}}{a_n}=\frac{\frac{(n!)^2(n+1)}{\left (2n+1\right )!(2n+3)}\cdot 4^{n}\cdot 2}{\frac{\left (n!\right )^2}{\left (2n+1\right )!}\cdot 4^n}=\frac{(n!)^2(n+1)\left (2n+1\right )!}{\left (2n+1\right )!(2n+3)(n!)^2}\cdot 2=\frac{(n+1)}{(2n+3)}\cdot 2=\frac{2n+2}{2n+3}\end{equation*}
It holds that $\displaystyle{\left |\frac{a_{n+1}}{a_n}\right |=\frac{2n+2}{2n+3}<1}$.
For the ratio test we have that the limit has to be less than $1$, right? So do we have to apply an other test?
- $\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}}$

Let $a_n=\prod_{i=1}^n\frac{2i-1}{2i}$ then $a_{n+1}=\prod_{i=1}^{n+1}\frac{2i-1}{2(i+1)}$.
So we get that $\frac{a_{n+1}}{a_n}=\frac{2(n+1)-1}{2(n+1+1)}=\frac{2n+1}{2n+4}<1$
Again the ration test doesn;t help us, does it?
- $\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n\cdot (2n+2)}}$

Let $a_n=\prod_{i=1}^n\frac{2i-1}{2i+2}$ then $a_{n+1}=\prod_{i=1}^{n+1}\frac{2i-1}{2i+2}$.
So we get that $\frac{a_{n+1}}{a_n}=\frac{2(n+1)-1}{2(n+1)+2}<1$.
The same here. So do we have to fins inequalities at each case or to use an other test? :unsure:

 

[I like Serena]

Hey mathmari!

Indeed, we cannot use the ratio test since its limit is 1.
I found the Raabe–Duhamel's test, which appears to work for all 3 of them.

 

[mathmari]

Indeed, we cannot use the ratio test since its limit is 1.
I found the Raabe–Duhamel's test, which appears to work for all 3 of them.

So do we have the following?

- $\frac{2n+2}{2n+3}=\frac{2n+3-1}{2n+3}=\frac{2n+3}{2n+3}-\frac{1}{2n+3}=1-\frac{1}{2n+3}$
$$1-\frac{1}{2n+3}\leq 1-\frac{b}{n} \Rightarrow \frac{b}{n}\leq \frac{1}{2n+3} \Rightarrow 2bn+3b\leq n \Rightarrow (1-2b)n\geq 3b \Rightarrow n\leq \frac{3b}{1-2b}$$

There is no $b>1$. So it diverges.
- $\frac{2n+1}{2n+4}=\frac{2n+4-3}{2n+4}=\frac{2n+4}{2n+4}-\frac{3}{2n+4}=1-\frac{3}{2n+4}$
$$1-\frac{3}{2n+4}\leq 1-\frac{b}{n} \Rightarrow \frac{b}{n}\leq \frac{3}{2n+4} \Rightarrow 2bn+4b\leq 3n \Rightarrow (3-2b)n\geq 4b$$ We choose $b=\frac{5}{4}$, then $n\geq 10$

So it converges.
- $\frac{a_{n+1}}{a_n}=\frac{2(n+1)-1}{2(n+1)+2}=\frac{2(n+1)+2-3}{2(n+1)+2}=\frac{2(n+1)+2}{2(n+1)+2}-\frac{3}{2(n+1)+2}=1-\frac{3}{2(n+1)+2}$
$$1-\frac{3}{2(n+1)+2}\leq 1-\frac{b}{n} \Rightarrow \frac{b}{n}\leq \frac{3}{2(n+1)+2}\Rightarrow \frac{b}{n}\leq \frac{3}{2n+4} \Rightarrow 2bn+4b\leq 3n \Rightarrow (3-2b)n\geq 4b$$ We choose $b=\frac{5}{4}$, then $n\geq 10$

So it converges. Is everything correct? :unsure:

 

[I like Serena]

So do we have the following?

- $\frac{2n+2}{2n+3}=\frac{2n+3-1}{2n+3}=\frac{2n+3}{2n+3}-\frac{1}{2n+3}=1-\frac{1}{2n+3}$
$$1-\frac{1}{2n+3}\leq 1-\frac{b}{n} \Rightarrow \frac{b}{n}\leq \frac{1}{2n+3} \Rightarrow 2bn+3b\leq n \Rightarrow (1-2b)n\geq 3b \Rightarrow n\leq \frac{3b}{1-2b}$$

There is no $b>1$. So it diverges.

What did you calculate?

You already had that $\frac{a_{n+1}}{a_n}=\frac{2n+2}{2n+3}$.
In the link that I posted, I found that we need to calculate $b_n=n\left(\frac{a_n}{a_{n+1}}-1\right)$.
So in this case we have $b_n=n\left(\frac{2n+3}{2n+2}-1\right)=\frac{n}{2n+2}\to \frac 12$.
Since the limit is $\frac 12<1$, it follows that the series indeed diverges. :geek:

- $\frac{2n+1}{2n+4}=\frac{2n+4-3}{2n+4}=\frac{2n+4}{2n+4}-\frac{3}{2n+4}=1-\frac{3}{2n+4}$
$$1-\frac{3}{2n+4}\leq 1-\frac{b}{n} \Rightarrow \frac{b}{n}\leq \frac{3}{2n+4} \Rightarrow 2bn+4b\leq 3n \Rightarrow (3-2b)n\geq 4b$$ We choose $b=\frac{5}{4}$, then $n\geq 10$

So it converges.

It seems you have $\frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+4}$, but I don't think that is correct.
I think there is a mistake in the post where you calculated it.
I found that it should be $\frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}$.

- $\frac{a_{n+1}}{a_n}=\frac{2(n+1)-1}{2(n+1)+2}=\frac{2(n+1)+2-3}{2(n+1)+2}=\frac{2(n+1)+2}{2(n+1)+2}-\frac{3}{2(n+1)+2}=1-\frac{3}{2(n+1)+2}$
$$1-\frac{3}{2(n+1)+2}\leq 1-\frac{b}{n} \Rightarrow \frac{b}{n}\leq \frac{3}{2(n+1)+2}\Rightarrow \frac{b}{n}\leq \frac{3}{2n+4} \Rightarrow 2bn+4b\leq 3n \Rightarrow (3-2b)n\geq 4b$$ We choose $b=\frac{5}{4}$, then $n\geq 10$

So it converges.

I don't know which calculation method you used. (Wondering)

Anyway, I calculated $b_n=n\left(\frac{a_n}{a_{n+1}}-1\right)=n\left(\frac{2n+4}{2n+1}-1\right) = \frac{3n}{2n+1}\to \frac 32$.
Since the limit is $\frac 32 >1$, it follows that the series indeed converges. :geek:

 

[mathmari]

Ahh I used the following part of your link:

An alternative formulation of this test is as follows. Let $\{ a_n \}$ be a series of real numbers. Then if b > 1 and K (a natural number) exist such that ${\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|\leq 1-{\frac {b}{n}}}$ for all n > K then the series {an} is convergent.

So is it better to use the other formulation? :unsure:

 

[I like Serena]

Ahh I used the following part of your link:

An alternative formulation of this test is as follows. Let $\{ a_n \}$ be a series of real numbers. Then if b > 1 and K (a natural number) exist such that ${\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|\leq 1-{\frac {b}{n}}}$ for all n > K then the series {an} is convergent.

So is it better to use the other formulation?

Aha. I actually did not read that far. :geek:
Now that I look at it, it seems indeed to me that the first formulation with $b_n$ is easier to apply and easier to understand and verify.

Either way, I just checked, and I can see now that your calculations and conclusions for the first and third problem are correct.
That leaves the second problem that I believe has a calculation mistake in $\frac{a_{n+1}}{a_n}$. Consequently the series diverges instead of converges.

 

[mathmari]

Aha. I actually did not read that far. :geek:
Now that I look at it, it seems indeed to me that the first formulation with $b_n$ is easier to apply and easier to understand and verify.

Either way, I just checked, and I can see now that your calculations and conclusions for the first and third problem are correct.
That leaves the second problem that I believe has a calculation mistake in $\frac{a_{n+1}}{a_n}$. Consequently the series diverges instead of converges.

Ahh yes!

So we have the following:

Let $\displaystyle{a_n=\prod_{i=1}^n\frac{2i-1}{2i}}$ .
Then $\displaystyle{a_{n+1}=\prod_{i=1}^{n+1}\frac{2i-1}{2i}}$.
So we get $\displaystyle{\left |\frac{a_{n+1}}{a_n}\right |=\frac{2(n+1)-1}{2(n+1)}=\frac{2n+2}{2n+2}-\frac{1}{2n+2}=1-\frac{1}{2n+2}}$
We have that \begin{equation*}1-\frac{1}{2n+2}\leq 1-\frac{b}{n} \Rightarrow \frac{b}{n}\leq \frac{1}{2n+2} \Rightarrow 2bn+2b\leq n \Rightarrow (1-2b)n\geq 2b \Rightarrow n\leq \frac{2b}{1-2b}\end{equation*} Thereisno such a $b>1$, so the series doesn't converge, right? :unsure:

 

[I like Serena]

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Great! Thank you!