Test comparison

**Apprentice123** · September 13th 2009, 06:15 PM

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

**redsoxfan325** · September 13th 2009, 06:32 PM

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.

**Apprentice123** · September 13th 2009, 06:51 PM

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

**Krizalid** · September 13th 2009, 07:26 PM

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 $n\ge1.$

**Apprentice123** · September 14th 2009, 12:49 PM

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 ?

**redsoxfan325** · September 14th 2009, 01:15 PM

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

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

Note that my choice of 1000 was arbitrary.

**Apprentice123** · September 14th 2009, 01:31 PM

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

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

Note that my choice of 1000 was arbitrary.

If not for all but finitely many

I can not say that converge?

**redsoxfan325** · September 14th 2009, 01:40 PM

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

**Apprentice123** · September 14th 2009, 02:56 PM

Originally Posted by redsoxfan325

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

Thank you.

This series converge?

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

**redsoxfan325** · September 14th 2009, 04:09 PM

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 $b_n=\frac{1}{n^2}$ .

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

**Apprentice123** · September 14th 2009, 05:40 PM

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

**redsoxfan325** · September 14th 2009, 05:44 PM

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.

**Apprentice123** · September 14th 2009, 05:48 PM

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

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