Infinite Series Test (Ratio Test get 1)

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  • #1
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Homework Statement


Test if the infinite series converge or diverge.

Homework Equations


[tex]
\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}
[/tex]


The Attempt at a Solution


I tried Ratio test:
[tex]a_{n+1} = \frac{4n+7}{(n+1)(n+2)(n+3)}[/tex]
[tex]a_{n} = \frac{4n+3}{n(n+1)(n+2)}[/tex]
[tex]\left|\frac{a_{n+1}}{a_{n}}\right| = \frac{4n+7}{(n+1)(n+2)(n+3)} \times \frac{n(n+1)(n+2)}{4n+3}
= \frac{n(4+7n)}{(n+3)(4n+3)}
= \frac{4n^{2}+7n}{4n^{2}+15n+9}[/tex]
[tex]lim_{n\rightarrow\infty} \left|\frac{a_{n+1}}{a_{n}}\right| = lim_{n\rightarrow\infty} \frac{4+\frac{7}{n}}{4+\frac{15}{n}+\frac{9}{n^{2}}} = \frac{4}{4} = 1[/tex]
The answer is inconclusive, and I can't seem to think of any other test yet.
Anyone can help me with this?
I will much appreciate it. Thanks!
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Think about what the terms look like for large n. Try a comparison test.
 
  • #3
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Think about what the terms look like for large n. Try a comparison test.
Thanks for your help.
This is what I done:
[tex]
n^{3} > n(n+1)(n+1)[/tex]
[tex]
\frac{4n+3}{n^{3}} < \frac{4n+2}{n(n+1)(n+2)}[/tex]
[tex]
\frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}}[/tex]
Both converge

Therefore, [tex]\sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)}[/tex] converges.


Is that correct?
 
Last edited:
  • #4
How is n^3 > n^3 + 2n^2 + n?
 
  • #5
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How is n^3 > n^3 + 2n^2 + n?
Isn't this the same with the [tex]

n^{3} > n(n+1)(n+1)
[/tex] above?
 
  • #6
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5,786
Isn't this the same with the [tex]

n^{3} > n(n+1)(n+1)
[/tex] above?
Yes, so why do you think that n3 > n(n + 1)(n + 2)?
 
  • #7
13
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Yes, so why do you think that n3 > n(n + 1)(n + 2)?
Opps, sorry,
found the careless mistake.

should be
n(n+1)(n+2) > n3

so
[tex]\frac{4n+3}{n(n+1)(n+2)} < \frac{4n+3}{n^{3}}[/tex]
Thanks for pointing me out

[tex]\frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}}[/tex]
Both converge

Therefore, [tex]\sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)}[/tex] converges.
 
  • #8
275
0
Go further with your inequalities:
(use the term by term comparison test)

[tex]
\frac{4n+3}{n(n+1)(n+2)}\leq \frac{7n}{n^{3}}= 7\frac{1}{n^{2}}
[/tex]
 
  • #9
13
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Go further with your inequalities:
(use the term by term comparison test)

[tex]
\frac{4n+3}{n(n+1)(n+2)}\leq \frac{7n}{n^{3}}= 7\frac{1}{n^{2}}
[/tex]
How do you get the 7n?
 
  • #11
13
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4n + 3 <= 7n for all n >= 1
Got it!
Thanks!
 
  • #12
13
0
So, the final answer should look like this:
[tex]
\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}
[/tex]

n(n+1)(n+2) [tex]\geq[/tex] n3
[tex]\frac{1}{n(n+1)(n+2)} \leq \frac{1}{n^{3}}[/tex]

4n+3 [tex]\leq[/tex] 7n , for all n [tex]\geq[/tex] 1

[tex]\frac{4n+3}{n(n+1)(n+2)} \leq \frac{7n}{n^{3}}[/tex]

[tex]\sum_{n=1}^{\infty}\frac{7n}{n^{3}} = \sum_{n=1}^{\infty}\frac{7}{n^{2}}[/tex]
Converge P-series (p > 1)

According to Comparison test,
since [tex]\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} \leq \sum_{n=1}^{\infty}\frac{7}{n^{2}}[/tex]
[tex]and \sum_{n=1}^{\infty}\frac{7}{n^{2}} converges[/tex],
[tex]therefore
\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} converges.
[/tex]
 
  • #13
Dick
Science Advisor
Homework Helper
26,260
619
So, the final answer should look like this:
[tex]
\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}
[/tex]

n(n+1)(n+2) [tex]\geq[/tex] n3
[tex]\frac{1}{n(n+1)(n+2)} \leq \frac{1}{n^{3}}[/tex]

4n+3 [tex]\leq[/tex] 7n , for all n [tex]\geq[/tex] 1

[tex]\frac{4n+3}{n(n+1)(n+2)} \leq \frac{7n}{n^{3}}[/tex]

[tex]\sum_{n=1}^{\infty}\frac{7n}{n^{3}} = \sum_{n=1}^{\infty}\frac{7}{n^{2}}[/tex]
Converge P-series (p > 1)

According to Comparison test,
since [tex]\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} \leq \sum_{n=1}^{\infty}\frac{7}{n^{2}}[/tex]
[tex]and \sum_{n=1}^{\infty}\frac{7}{n^{2}} converges[/tex],
[tex]therefore
\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} converges.
[/tex]
Very nice.
 
  • #14
13
0
Very nice.
Thanks,
and thanks for everyone that helps.
:biggrin:
 

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