Thread: Test comparison

Resolving a list of exercise these two exercises I could not

Determined using the comparison test

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

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

Originally Posted by Apprentice123

Resolving a list of exercise these two exercises I could not

Determined using the comparison test

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

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

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

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

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

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

Thank you very much

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

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

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

As long as $\displaystyle n\ge1.$

Originally Posted by Krizalid

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

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

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

As long as $\displaystyle n\ge1.$

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

Originally Posted by Apprentice123

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

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

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

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

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

Note that my choice of 1000 was arbitrary.

Originally Posted by redsoxfan325

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

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

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

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

Note that my choice of 1000 was arbitrary.

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

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

Originally Posted by redsoxfan325

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

Thank you.


This series converge?


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

Originally Posted by Apprentice123

Thank you.


This series converge?


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

Try the limit comparison test with $\displaystyle b_n=\frac{1}{n^2}$.

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

Originally Posted by redsoxfan325

Try the limit comparison test with $\displaystyle b_n=\frac{1}{n^2}$.

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

This limit is not a number

Originally Posted by Apprentice123

This limit is not a number

$\displaystyle \lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = \lim_{n\to\infty}\frac{n}{\sqrt{n^2-1}} = \lim_{n\to\infty}\frac{n}{n\sqrt{1-1/n^2}} = \lim_{n\to\infty}\frac{1}{\sqrt{1-1/n^2}}=\frac{1}{\sqrt{1-0}}=1$

So $\displaystyle a_n$ converges because $\displaystyle b_n$ does.

Originally Posted by redsoxfan325

$\displaystyle \lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = \lim_{n\to\infty}\frac{n}{\sqrt{n^2-1}} = \lim_{n\to\infty}\frac{n}{n\sqrt{1-1/n^2}} = \lim_{n\to\infty}\frac{1}{\sqrt{1-1/n^2}}=\frac{1}{\sqrt{1-0}}=1$

So $\displaystyle a_n$ converges because $\displaystyle b_n$ does.

Thank you