# Convergence of a series

#### chwala

Gold Member
1. The problem statement, all variables and given/known data
Determine whether the series $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ converges or not

2. Relevant equations

3. 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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#### tnich

Homework Helper
1. The problem statement, all variables and given/known data
Determine whether the series $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ converges or not

2. Relevant equations

3. 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

Science Advisor
Homework Helper
Dearly Missed
1. The problem statement, all variables and given/known data
Determine whether the series $\frac {(n^3+3n)^{1/2}} {5n^3+3n^2+2 sin (n)}$ converges or not

2. Relevant equations

3. 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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#### chwala

Gold Member
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

Gold Member
there look at my working attached........

#### Attachments

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#### chwala

Gold Member
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

Science Advisor
Homework Helper
Dearly Missed
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 Gold Member 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

Gold Member
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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#### Delta2

Homework Helper
Gold Member
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."

#### chwala

Gold Member
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...

#### Mark44

Mentor
@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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#### Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
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

Gold Member
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

Gold Member
@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

Mentor
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

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

#### chwala

Gold Member
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

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

#### SammyS

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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

Mentor
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

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

#### chwala

Gold Member
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

Mentor
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?

#### chwala

Gold Member
ok let me look at this using comparison test

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