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

• chwala
In summary: 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.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 chwala Gold Member ## 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,... chwala said: ## 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? chwala said: ## 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).## Last edited: Delta2 and chwala Ray Vickson said: 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.. there look at my working attached... #### Attachments • convergenceproblem.docx 14.9 KB · Views: 264 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 chwala said: 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
Ray Vickson said:
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...

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

Last edited:
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
Delta² said:
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...

Delta2
@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}

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chwala said:
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
Ray Vickson said:
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...

Mark44 said:
@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...

Last edited:
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...

Mark44 said:
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.

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

chwala said:
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.

member 587159
chwala said:
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.

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

Mark44 said:
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...

From post #9, of July 18:
chwala said:
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?

valenumr
ok let me look at this using comparison test

chwala said:
ok let me look at this using comparison test
The Limit Comparison Test would be the better, and easier choice.

Mark44 said:
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

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.

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chwala said:
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
Mark44 said:
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

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## 1. Does the series converge or diverge?

The answer to this question depends on the value of n. If n is a positive integer, the series will converge. However, if n is a negative integer or a non-integer value, the series will diverge.

## 2. How do you determine if the series converges?

To determine if the series converges, we can use the ratio test or the root test. If the limit of the absolute value of the ratio or the root is less than 1, the series will converge. Otherwise, it will diverge.

## 3. Can you simplify the series to make it easier to determine convergence?

Yes, we can simplify the series by factoring out the highest power of n in the numerator and denominator. This will help us to apply the ratio or root test more easily.

## 4. Can you use the comparison test to determine convergence?

No, the comparison test cannot be used for this series since there is no other series with known convergence properties that is larger or smaller than this series.

## 5. Can you use the integral test to determine convergence?

Yes, the integral test can be used to determine the convergence of this series. By finding the antiderivative of the series, we can compare it to the integral of 1/n, which is a known convergent series. If the antiderivative is also convergent, then the original series will also converge.

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