Does the Series ## \frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}## Converge?

[chwala]

Homework Statement

Determine whether the series $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ converges or not

Homework Equations

The Attempt at a Solution

looking at $1/sin (n)$ by comparison,
$1/n^2=1+1/4+1/9+1/16+...$ converges for $n≥1$
for $n≥1$
implying that ${sin (n)}≤n$
$1/2sin (n) ≤1/{n}$ converges, up to this point i have been trying to look at the trig. term in the series...
again by comparison,...

 

[tnich]

Homework Statement

Determine whether the series $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ converges or not

Homework Equations

The Attempt at a Solution

looking at $1/sin (n)$ by comparison,
$1/n^2=1+1/4+1/9+1/16+...$ converges for $n≥1$
for $n≥1$
implying that ${sin (n)}≤n$
$1/2sin (n) ≤1/{n}$ converges, up to this point i have been trying to look at the trig. term in the series...
again by comparison,...

I think the first thing you need to do is eliminate the trig term by finding a bounding series. What series $f(n)$ can you come up that is just a little bit larger than $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ and has no trig term?

 

[Ray Vickson]

Homework Statement

Determine whether the series $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ converges or not

Homework Equations

The Attempt at a Solution

looking at $1/sin (n)$ by comparison,
$1/n^2=1+1/4+1/9+1/16+...$ converges for $n≥1$
for $n≥1$
implying that ${sin (n)}≤n$
$1/2sin (n) ≤1/{n}$ converges, up to this point i have been trying to look at the trig. term in the series...
again by comparison,...

Look at
$$\text{numerator} = (n^3 +3n)^{1/2} = n^{3/2} \left( 1 + \frac 3 n \right)^{1/2} $$
and
$$\text{denominator} = 5n^3 + 3 n^2 + 2 \sin(n) = 5 n^3 \left( 1 + \frac{3}{5n} + \frac{2 \sin(n)}{5n^3} \right).$$
The numerator is $< 2 \, n^{3/2}$, and for any small $\epsilon > 0$ we can find $N> 0$ so that $n > N$ implies the denominator is $> 5n^3 (1 - \epsilon).$

 

[chwala]

Look at
$$\text{numerator} = (n^3 +3n)^{1/2} = n^{3/2} \left( 1 + \frac 3 n \right)^{1/2} $$
and
$$\text{denominator} = 5n^3 + 3 n^2 + 2 \sin(n) = 5 n^3 \left( 1 + \frac{3}{5n} + \frac{2 \sin(n)}{5n^3} \right).$$
The numerator is $< 2 \, n^{3/2}$, and for any small $\epsilon > 0$ we can find $N> 0$ so that $n > N$ implies the denominator is $> 5n^3 (1 - \epsilon).$

let me post my attempt..

 

[chwala]

there look at my working attached...

 

[chwala]

since the series $1/n^3$ converges for the p-series p greater than 1, it suffices to say that our series converges though it fails for the limit comparison test...my thoughts

 

[Ray Vickson]

there look at my working attached...

This attachment shows up on my laptop computer, but opens empty and blank on my i-phone. You should avoid using tools that require the reader to use particular media to read your posts. Avoid WORD; use LaTeX instead. It is faster, easier and looks better. Now is the time for you to start learning it; you can install a complete system on your computer for $0 (for typing up your assignments), and it is built-in in this Forum as well.

 

[chwala]

This attachment shows up on my laptop computer, but opens empty and blank on my i-phone. You should avoid using tools that require the reader to use particular media to read your posts. Avoid WORD; use LaTeX instead. It is faster, easier and looks better. Now is the time for you to start learning it; you can install a complete system on your computer for $0 (for typing up your assignments), and it is built-in in this Forum as well.

ok sir, allow me to repost in latex...doing so in a few hours let me finish with a class...

 

[chwala]

our problem $\sum_{n=1}^\infty$ $\frac {(n^2+3n)^{1/2}} {5n^3+3n^2+ 2 sin n}\$
*give me a few more hours finishing with a class*
$sum_{n=1}^\infty$ ${\frac {3}{n^2}+1}$
GIVE ME TIMEAM IN CLASS

 

[Delta2]

I also can't read your work in word document (yeap living in 2018 and haven't installed Microsoft Word :D)…

But , following the hints from @Ray Vickson, the final sequence $b_n$ which we ll use for the direct comparison test is $b_n=\frac{2n^{3/2}}{5\epsilon n^3}=\frac{2}{5\epsilon}\frac{1}{n^{3/2}}$ so your final conclusion should be "because the series of $b_n$ converges (p-series with p=3/2>1) and $|a_n|<|b_n|$ for n>N for some positive $\epsilon$ and $N$, from direct comparison test it follows that the series of $a_n$ converges."

 

[chwala]

I also can't read your work in word document (yeap living in 2018 and haven't installed Microsoft Word :D)…

But , following the hints from @Ray Vickson, the final sequence $b_n$ which we ll use for the direct comparison test is $b_n=\frac{2n^{3/2}}{5\epsilon n^3}=\frac{2}{5\epsilon}\frac{1}{n^{3/2}}$ so your final conclusion should be "because the series of $b_n$ converges (p-series with p=3/2>1) and $|a_n|<|b_n|$ for n>N for some positive $\epsilon$ and $N$, from direct comparison test it follows that the series of $a_n$ converges."

kindly let's be patient as i am reposting the solution using latex, i am learning a few latex terms...give me time...

 

[Mark44]

@chwala, the work you showed in the Word document is wrong right from the start.
You have
$$\sum_{n = 1}^\infty \frac{\sqrt{n^2 + 3n}}{5n^3 + 3n^2 + 2\sin n} = \sum_{n = 1}^\infty \frac {\frac{\sqrt{n(n^2 + 3)}}{n^3}}{5 + \frac 3 n + \frac{2\sin n}{n^3}}$$

In post #1 you have $(n^3 + 3n)^{1/2}$ in the numerator, but above you have $\sqrt{n^2 + 3n}$ on the left side. That's a relatively minor typo, but confusing for someone who might not have read things carefully throughout the thread.

BTW, since you're struggling with LaTeX, here's what I typed for the stuff above, in unrendered form:
\sum_{n = 1}^\infty \frac{\sqrt{n^2 + 3n}}{5n^3 + 3n^2 + 2\sin n} = \sum_{n = 1}^\infty \frac {\frac{\sqrt{n(n^2 + 3)}}{n^3}}{5 + \frac 3 n + \frac{2\sin n}{n^3}}

One thing to keep in mind for exponents, subscripts, fraction parts, limits of integration, and possibly a few more things: if the exponent, fraction part, etc. is just a single character, you don't need to include braces around it.
For example, this is fine: x^2
But here you need braces: x^{-2}

And this is fine \frac 1 2
But here you need braces: \frac {x - 2} 3 and \frac {x - 2}{3x}

 

[Ray Vickson]

kindly let's be patient as i am reposting the solution using latex, i am learning a few latex terms...give me time...

Nobody is rushing you. If you need a few days, take a few days.

 

[chwala]

Nobody is rushing you. If you need a few days, take a few days.

thanks i have been busy now i can embark on physicsforums...

 

[chwala]

@chwala, the work you showed in the Word document is wrong right from the start.
You have
$$\sum_{n = 1}^\infty \frac{\sqrt{n^2 + 3n}}{5n^3 + 3n^2 + 2\sin n} = \sum_{n = 1}^\infty \frac {\frac{\sqrt{n(n^2 + 3)}}{n^3}}{5 + \frac 3 n + \frac{2\sin n}{n^3}}$$

In post #1 you have $(n^3 + 3n)^{1/2}$ in the numerator, but above you have $\sqrt{n^2 + 3n}$ on the left side. That's a relatively minor typo, but confusing for someone who might not have read things carefully throughout the thread.

BTW, since you're struggling with LaTeX, here's what I typed for the stuff above, in unrendered form:
\sum_{n = 1}^\infty \frac{\sqrt{n^2 + 3n}}{5n^3 + 3n^2 + 2\sin n} = \sum_{n = 1}^\infty \frac {\frac{\sqrt{n(n^2 + 3)}}{n^3}}{5 + \frac 3 n + \frac{2\sin n}{n^3}}

One thing to keep in mind for exponents, subscripts, fraction parts, limits of integration, and possibly a few more things: if the exponent, fraction part, etc. is just a single character, you don't need to include braces around it.
For example, this is fine: x^2
But here you need braces: x^{-2}

And this is fine \frac 1 2
But here you need braces: \frac {x - 2} 3 and \frac {x - 2}{3x}

noted, let me retype my whole solution using latex...

 

[Mark44]

While you're figuring out LaTeX, something you should consider is that
$$\sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2} \le \sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2\sin(n)} \le \sum_{n = 1}^\infty\frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 - 2}$$
If you can show that the outer summations are convergent, you'll be able to say something about the series you're working with.

 

[chwala]

am back guys...was on vacation, let me embark on this again...

 

[chwala]

While you're figuring out LaTeX, something you should consider is that
$$\sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2} \le \sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2\sin(n)} \le \sum_{n = 1}^\infty\frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 - 2}$$
If you can show that the outer summations are convergent, you'll be able to say something about the series you're working with.

bet its easy to show by comparison method.

 

[chwala]

Allow me to look at this post again, i hope its not closed. I have been busy lately...

 

[SammyS]

Allow me to look at this post again, i hope its not closed. I have been busy lately...

It looks to me like it's open.

 

[Mark44]

Allow me to look at this post again, i hope its not closed. I have been busy lately...

Yes, it's open, although 5+ months seems a long time to figure out this problem.

 

[chwala]

Let me look at this again, i need to refresh on convergence, my apologies

 

[chwala]

Yes, it's open, although 5+ months seems a long time to figure out this problem.

can you give me more insight Mark? i am stuck in this rumble...

 

[Mark44]

From post #9, of July 18:

our problem $\sum_{n=1}^\infty$ $\frac {(n^2+3n)^{1/2}} {5n^3+3n^2+ 2 sin n}\$

The dominant term in the numerator is n, and the dominant term in the denominator is $5n^3$. Do you know the Limit Comparison Test?

 

[chwala]

ok let me look at this using comparison test

 

[Mark44]

ok let me look at this using comparison test

The Limit Comparison Test would be the better, and easier choice.

 

[chwala]

While you're figuring out LaTeX, something you should consider is that
$$\sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2} \le \sum_{n = 1}^\infty \frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 + 2\sin(n)} \le \sum_{n = 1}^\infty\frac{(n^3 + 3n)^{1/2}}{5n^3 + 3n^2 - 2}$$
If you can show that the outer summations are convergent, you'll be able to say something about the series you're working with.

Mark i want to respond to this thread...its bad of me to not have given it priority...i will respond over the weekend...cheers

 

[chwala]

From Post $3$ and $10$, we make use of the comparison test by making use of another similar series to determine whether our series is convergent or not.
We have,
$\dfrac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$=$\dfrac {n^3(1+\frac{3}{n})^{1/2}} {5n^3(1+\frac{3}{5n}+\frac{2 sin (n)}{5n^3})}$≤$\dfrac {2n^{\frac {3}{2}} }{5εn^3}$=$\dfrac {2}{5εn^{\frac {3}{2}}}$
Using $p$ series, the series converges because $\dfrac{3}{2}$>$1$. This Implies that our series is convergent.

 

[Mark44]

From Post $3$ and $10$, we make use of the comparison test by making use of another similar series to determine whether our series is convergent or not.
We have,
$\dfrac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$=$\dfrac {n^3(1+\frac{3}{n})^{1/2}} {5n^3(1+\frac{3}{5n}+\frac{2 sin (n)}{5n^3})}$≤$\dfrac {2n^{\frac {3}{2}} }{5εn^3}$=$\dfrac {2}{5εn^{\frac {3}{2}}}$
Using $p$ series, the series converges because $\dfrac{3}{2}$>$1$. This Implies that our series is convergent.

1. Is that $\epsilon$ in the 3rd and 4th expressions a typo? I don't see why it should be there.
2. There's a typo in the 2nd expression. The $n^3$ that you brought out should be $n^{3/2}$.
3. You're on somewhat shaky ground in saying that the 2nd expression is $\le$ the third expression. If the numerator in the 2nd expression had been only $n^3$ with the same denominator it would be clear that 2nd expression $\le$ your third expression. However, when both numerator and denominator are increasing, it's more difficult to make that case. Instead of the comparison test, I would use the limit comparison test.

For a simpler example of what I'm talking about in #3, consider $\frac{1 + x}{3 + y}$ vs. $\frac 2 3$. Is the first fraction smaller or larger than the second? It's clear that $\frac 1 {3 + y} \le \frac 1 3$ for positive y, but having both numerator and denominator changing makes it more difficult to prove.

 

[chwala]

1. Is that $\epsilon$ in the 3rd and 4th expressions a typo? I don't see why it should be there.
2. There's a typo in the 2nd expression. The $n^3$ that you brought out should be $n^{3/2}$.
3. You're on somewhat shaky ground in saying that the 2nd expression is $\le$ the third expression. If the numerator in the 2nd expression had been only $n^3$ with the same denominator it would be clear that 2nd expression $\le$ your third expression. However, when both numerator and denominator are increasing, it's more difficult to make that case. Instead of the comparison test, I would use the limit comparison test.

For a simpler example of what I'm talking about in #3, consider $\frac{1 + x}{3 + y}$ vs. $\frac 2 3$. Is the first fraction smaller or larger than the second? It's clear that $\frac 1 {3 + y} \le \frac 1 3$ for positive y, but having both numerator and denominator changing makes it more difficult to prove.

Thanks Mark, I will use limits and further spend the next two days refreshing on convergence/divergence of sequences and series...cheers...It was epsilon...I will amend the expression.

Mark by the way just clarify on this, I am making my remarks based on my other post on convergence...and I had used the understanding of limits to conclude convergence...but was informed its not right approach.
Can one make use of limits in finding convergence of a series? I thought limits are applicable to sequences only...cheers