1. ## Test comparison

Resolving a list of exercise these two exercises I could not

Determined using the comparison test

1) $\sum_{n=1}^{\infty} \frac{(n+1)}{n^3}$

2) $\sum_{n=0}^{\infty} \frac{n^2+1}{n^3+1}$

2. Originally Posted by Apprentice123
Resolving a list of exercise these two exercises I could not

Determined using the comparison test

1) $\sum_{n=1}^{\infty} \frac{(n+1)}{n^3}$

2) $\sum_{n=0}^{\infty} \frac{n^2+1}{n^3+1}$
The limit comparison test says that:

• If the limit of a[n]/b[n] is positive, then the sum of a[n] converges if and only if the sum of b[n] converges.
• If the limit of a[n]/b[n] is zero, and the sum of b[n] converges, then the sum of a[n] also converges.
• If the limit of a[n]/b[n] is infinite, and the sum of b[n] diverges, then the sum of a[n] also diverges.

For the first one, compare it to $\frac{1}{n^2}$. $\lim_{n\to\infty}\frac{\frac{(n+1)}{n^3}}{\frac{1} {n^2}} = 1$, so the sum converges because $\sum_{n=1}^{\infty}\frac{1}{n^2}$ does.

For the second one, compare it to $\frac{1}{n+1}$. $\lim_{n\to\infty}\frac{\frac{n^2+1}{n^3+1}}{\frac{ 1}{n+1}}=1$, so the sum diverges because $\sum_{n=0}^{\infty}\frac{1}{n+1}$ does.

3. Originally Posted by redsoxfan325
The limit comparison test says that:

• If the limit of a[n]/b[n] is positive, then the sum of a[n] converges if and only if the sum of b[n] converges.
• If the limit of a[n]/b[n] is zero, and the sum of b[n] converges, then the sum of a[n] also converges.
• If the limit of a[n]/b[n] is infinite, and the sum of b[n] diverges, then the sum of a[n] also diverges.

For the first one, compare it to $\frac{1}{n^2}$. $\lim_{n\to\infty}\frac{\frac{(n+1)}{n^3}}{\frac{1} {n^2}} = 1$, so the sum converges because $\sum_{n=1}^{\infty}\frac{1}{n^2}$ does.

For the second one, compare it to $\frac{1}{n+1}$. $\lim_{n\to\infty}\frac{\frac{n^2+1}{n^3+1}}{\frac{ 1}{n+1}}=1$, so the sum diverges because $\sum_{n=0}^{\infty}\frac{1}{n+1}$ does.

Thank you very much

4. i read comparison test, so we can actually use the direct comparison test.

1) $\frac{n+1}{n^{3}}\le \frac{2n}{n^{3}}=\frac{2}{n^{2}}.$

2) $\frac{n^{2}+1}{n^{3}+1}>\frac{n^{2}}{2n^{3}}=\frac {1}{2n}.$

As long as $n\ge1.$

5. Originally Posted by Krizalid
i read comparison test, so we can actually use the direct comparison test.

1) $\frac{n+1}{n^{3}}\le \frac{2n}{n^{3}}=\frac{2}{n^{2}}.$

2) $\frac{n^{2}+1}{n^{3}+1}>\frac{n^{2}}{2n^{3}}=\frac {1}{2n}.$

As long as $n\ge1.$
I have a doubt.
If I have one series $a_n$ I want to determine the convergence, found that the series $b_n$ is convergent, but $b_n$ > $a_n$ up to a value of n, not for all values of n. The series $a_n$ is convergent ?

6. Originally Posted by Apprentice123
I have a doubt.
If I have one series $a_n$ I want to determine the convergence, found that the series $b_n$ is convergent, but $b_n$ > $a_n$ up to a value of n, not for all values of n. The series $a_n$ is convergent ?
Not necessarily. If you are comparing $a_n$ to $b_n$ and you know that $b_n$ converges, $a_n\leq b_n$ for all but finitely many $n$ will tell you that $a_n$ converges.

For example, say $b_n$ converges. You compare $a_n$ to $b_n$ and find:

1.) $\left\{\begin{array}{lr}a_nb_n &: n>1000\end{array}\right\}$ $a_n$ does not necessarily converge.

2.) $\left\{\begin{array}{lr}a_n>b_n &: n\leq 1000\\a_n1000\end{array}\right\}$ $a_n$ will converge.

Note that my choice of 1000 was arbitrary.

7. Originally Posted by redsoxfan325
Not necessarily. If you are comparing $a_n$ to $b_n$ and you know that $b_n$ converges, $a_n\leq b_n$ for all but finitely many $n$ will tell you that $a_n$ converges.

For example, say $b_n$ converges. You compare $a_n$ to $b_n$ and find:

1.) $\left\{\begin{array}{lr}a_nb_n &: n>1000\end{array}\right\}$ $a_n$ does not necessarily converge.

2.) $\left\{\begin{array}{lr}a_n>b_n &: n\leq 1000\\a_n1000\end{array}\right\}$ $a_n$ will converge.

Note that my choice of 1000 was arbitrary.

If not for all but finitely many I can not say that converge?

8. No. It could possibly converge, but the comparison test does not prove it.

9. Originally Posted by redsoxfan325
No. It could possibly converge, but the comparison test does not prove it.
Thank you.

This series converge?

$\sum_{n=2}^{\infty} \frac{1}{n\sqrt{n^2-1}}$

10. Originally Posted by Apprentice123
Thank you.

This series converge?

$\sum_{n=2}^{\infty} \frac{1}{n\sqrt{n^2-1}}$
Try the limit comparison test with $b_n=\frac{1}{n^2}$.

$\lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = ???$

11. Originally Posted by redsoxfan325
Try the limit comparison test with $b_n=\frac{1}{n^2}$.

$\lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = ???$
This limit is not a number

12. Originally Posted by Apprentice123
This limit is not a number
$\lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = \lim_{n\to\infty}\frac{n}{\sqrt{n^2-1}} = \lim_{n\to\infty}\frac{n}{n\sqrt{1-1/n^2}} = \lim_{n\to\infty}\frac{1}{\sqrt{1-1/n^2}}=\frac{1}{\sqrt{1-0}}=1$

So $a_n$ converges because $b_n$ does.

13. Originally Posted by redsoxfan325
$\lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = \lim_{n\to\infty}\frac{n}{\sqrt{n^2-1}} = \lim_{n\to\infty}\frac{n}{n\sqrt{1-1/n^2}} = \lim_{n\to\infty}\frac{1}{\sqrt{1-1/n^2}}=\frac{1}{\sqrt{1-0}}=1$

So $a_n$ converges because $b_n$ does.
Thank you