# Limit Comparision Test: Get Help Now

• MHB
• jaychay
In summary: The conclusion of the implication is that if $\lim a_k/b_k=c$ and $0<c<\infty$, then the answer for $\sum b_k$ is the same as the one for $\sum a_k$.
jaychay

Have you tried to apply the limit comparison test or find the limit of the terms as $k\to\infty$?

Evgeny.Makarov said:
Have you tried to apply the limit comparison test or find the limit of the terms as $k\to\infty$?
okay I will try to do it from what you guide me and I will let you check it for me.

Evgeny.Makarov said:
Have you tried to apply the limit comparison test or find the limit of the terms as $k\to\infty$?

I am struggle at this point sir

Why didn't you include the numerator $k$ in $a_k$ and $b_k$? Your $b_k$ is different from the series in the problem statement.

I suggest using $a_k=1/k^2$ and $b_k$ as in the original series. Also use the following rule. If $f(x)=ax^m+\sum_{i=0}^{m-1}a_ix^i$ and $g(x)=bx^n+\sum_{i=0}^{n-1}b_ix^i$, then
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\begin{cases}0,&m<n\\a/b,&m=n.\\\infty,&m>n\end{cases}$$

Evgeny.Makarov said:
Why didn't you include the numerator $k$ in $a_k$ and $b_k$? Your $b_k$ is different from the series in the problem statement.

I suggest using $a_k=1/k^2$ and $b_k$ as in the original series. Also use the following rule. If $f(x)=ax^m+\sum_{i=0}^{m-1}a_ix^i$ and $g(x)=bx^n+\sum_{i=0}^{n-1}b_ix^i$, then
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\begin{cases}0,&m<n\\a/b,&m=n.\\\infty,&m>n\end{cases}$$
Can you guide me more because I am very confused right now ?

The limit comparison test says that in order to ascertain convergence of $\sum b_k$ one can come up with another series $\sum a_k$ for which it is known whether it converges. If $\lim a_k/b_k=c$ and $0<c<\infty$, then the answer for $\sum b_k$ is the same as the one for $\sum a_k$.

I suggest considering $\sum b_k$ to be the series from the problem statement and $a_k=1/k^2$. It is known that $\sum 1/k^2$ converges. What limit do we have to consider to determine the convergence for $\sum b_k$?

jaychay said:

note ...

$\dfrac{k}{k+2k^2+3k^3} = \dfrac{1}{1+2k+3k^2} < \dfrac{1}{k^2} \text{ for all } k > 1$

$\displaystyle \implies \sum_{k=1}^\infty \dfrac{1}{1+2k+3k^2} \text{ converges by direct comparison to the known convergent series } \sum_{k=1}^\infty \dfrac{1}{k^2}$

Evgeny.Makarov said:
The limit comparison test says that in order to ascertain convergence of $\sum b_k$ one can come up with another series $\sum a_k$ for which it is known whether it converges. If $\lim a_k/b_k=c$ and $0<c<\infty$, then the answer for $\sum b_k$ is the same as the one for $\sum a_k$.

I suggest considering $\sum b_k$ to be the series from the problem statement and $a_k=1/k^2$. It is known that $\sum 1/k^2$ converges. What limit do we have to consider to determine the convergence for $\sum b_k$?
So here is the work that I have done recently so how can I determine that it is convergent or divergent ?

#### Attachments

• IMG_20210824_0001.jpg
406.9 KB · Views: 98
jaychay said:
so how can I determine that it is convergent or divergent ?
Evgeny.Makarov said:
It is known that $\sum 1/k^2$ converges.
In fact, $\sum_{k=1}^\infty 1/k^2=\pi^2/6$ (see Wikipedia).

Evgeny.Makarov said:
In fact, $\sum_{k=1}^\infty 1/k^2=\pi^2/6$ (see Wikipedia).
I still do not understand what you are trying to tell me sir.

The limit comparison test is a theorem that has the following form: "For every two series $\sum a_k$ and $\sum b_k$, if ..., then ...". What is the conclusion of this implication, i.e., what is written after "then"?

## 1. What is the Limit Comparison Test?

The Limit Comparison Test is a method used in calculus to determine the convergence or divergence of a series. It involves comparing the given series to a known convergent or divergent series to determine its behavior.

## 2. When should the Limit Comparison Test be used?

The Limit Comparison Test should be used when the terms of the given series are positive and the series is not a geometric series. It is also useful when the series contains factorial or exponential terms.

## 3. How is the Limit Comparison Test performed?

The Limit Comparison Test involves taking the limit of the ratio between the terms of the given series and the terms of the known series. If the limit is a finite positive number, then the two series have the same behavior. If the limit is 0 or infinite, then the given series will have the opposite behavior of the known series.

## 4. What is the difference between the Limit Comparison Test and the Ratio Test?

The Limit Comparison Test and the Ratio Test are both used to determine the convergence or divergence of a series. However, the Limit Comparison Test compares the given series to a known series, while the Ratio Test compares the given series to itself. Additionally, the Limit Comparison Test is often easier to use for more complicated series.

## 5. What are the limitations of the Limit Comparison Test?

The Limit Comparison Test can only be used on series with positive terms. It also requires the existence of a known series with known convergence or divergence. If the limit of the ratio is inconclusive, other tests may need to be used to determine the behavior of the series.

Replies
6
Views
1K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
12
Views
1K
Replies
2
Views
2K
Replies
6
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
4
Views
2K