# Checking Convergence of Series: Inequalities & Tests

• MHB
• mathmari
In summary, we discussed the convergence of three series using the Raabe-Duhamel's test. In the first series, the limit was equal to 1, so we could not use the ratio test. However, we found that the series diverges by showing that the limit of $b_n=n\left(\frac{a_n}{a_{n+1}}-1\right)$ is greater than 1. For the second and third series, the limit was less than 1, so by the Raabe-Duhamel's test, we found that the second series diverges and the third series converges. We also discussed the alternative formulation of the test and concluded that it may be easier to apply and understand.
mathmari
Gold Member
MHB
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:

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.

Klaas van Aarsen said:
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:

mathmari said:
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:
mathmari said:
- $\frac{2n+1}{2n+4}=\frac{2n+4-3}{2n+4}=\frac{2n+4}{2n+4}-\frac{3}{2n+4}=1-\frac{3}{2n+4}$
$$1-\frac{3}{2n+4}\leq 1-\frac{b}{n} \Rightarrow \frac{b}{n}\leq \frac{3}{2n+4} \Rightarrow 2bn+4b\leq 3n \Rightarrow (3-2b)n\geq 4b$$ 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 $\frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}$.
mathmari said:
- $\frac{a_{n+1}}{a_n}=\frac{2(n+1)-1}{2(n+1)+2}=\frac{2(n+1)+2-3}{2(n+1)+2}=\frac{2(n+1)+2}{2(n+1)+2}-\frac{3}{2(n+1)+2}=1-\frac{3}{2(n+1)+2}$
$$1-\frac{3}{2(n+1)+2}\leq 1-\frac{b}{n} \Rightarrow \frac{b}{n}\leq \frac{3}{2(n+1)+2}\Rightarrow \frac{b}{n}\leq \frac{3}{2n+4} \Rightarrow 2bn+4b\leq 3n \Rightarrow (3-2b)n\geq 4b$$ 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:

Last edited:

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:

mathmari said:

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.

Klaas van Aarsen said:
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:

Yep. (Nod)

Klaas van Aarsen said:
Yep. (Nod)

Great! Thank you!

## 1. What is the definition of convergence for a series?

The definition of convergence for a series is when the sum of all its terms approaches a finite value as the number of terms increases.

## 2. How do you determine if a series is convergent or divergent?

To determine if a series is convergent or divergent, you can use various tests such as the comparison test, ratio test, or root test. These tests compare the given series to a known series with known convergence or divergence properties.

## 3. What is the difference between absolute convergence and conditional convergence?

Absolute convergence is when a series converges regardless of the order of its terms, while conditional convergence is when a series only converges if its terms are arranged in a specific order. In other words, absolute convergence is more strict than conditional convergence.

## 4. Can a series be both convergent and divergent?

No, a series cannot be both convergent and divergent. It can only be one or the other. If a series does not converge, it is automatically considered divergent.

## 5. How do you use inequalities to check for convergence of a series?

Inequalities can be used to compare the given series to a known series with known convergence properties. This can help determine if the given series is convergent or divergent. For example, if the given series is always less than or equal to a convergent series, then the given series must also be convergent.

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