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Convergence of a series

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  • #1
chwala
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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,...
 

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  • #2
tnich
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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?
 
  • #3
Ray Vickson
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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).##
 
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  • #4
chwala
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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..
 
  • #5
chwala
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there look at my working attached........
 

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  • #6
chwala
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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
 
  • #7
Ray Vickson
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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.
 
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  • #8
chwala
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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...
 
  • #9
chwala
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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
 
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  • #10
Delta2
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I also cant 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."
 
  • #11
chwala
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I also cant 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...
 
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  • #12
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@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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  • #13
Ray Vickson
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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.
 
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  • #14
chwala
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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...
 
  • #15
chwala
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@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...
 
  • #16
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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.
 
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  • #17
chwala
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am back guys...was on vacation, let me embark on this again...
 
  • #18
chwala
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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.
 
  • #19
chwala
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Allow me to look at this post again, i hope its not closed. I have been busy lately...
 
  • #20
SammyS
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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.
 
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  • #21
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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.
 
  • #22
chwala
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Let me look at this again, i need to refresh on convergence, my apologies
 
  • #23
chwala
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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...
 
  • #24
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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?
 
  • #25
chwala
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ok let me look at this using comparison test :smile:
 

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