A succession with a serie

  • #1
Hi people...

I've tried to solve the limit of this succession when n goes to inf, but it's been quite hard...

the general term of the succesion 'n' appears both on the top limit of the serie and also inside the general term of the serie...

I attach the problem

My problem is, if this term 'n' appears both on the top limit of the serie and also inside the serie, how do I solve the limit as n->Inf...

What steps shall I take...

Many thanks
 

Attachments

  • série.jpg
    série.jpg
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Answers and Replies

  • #2
Sorry it is not a succession, it is a sequence!!! Bad translation from Portuguese :)
How do I solve the limit of this sequence?

Many thanks for your attention
 
  • #3
mathman
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General term in the sum is n2/(5n3 + 2k). We can bracket it as follows:
n2/(5n3)>n2/(5n3 + 2k)>n2/(5n3 + 2n)
Both brackets are independent of k and there are n+1 terms.
Therefore the left bracket sums to (n+1)n2/(5n3) -> 1/5.

The right hand bracket sums to (n+1)n2/(5n3 + 2n)-> 1/5.

Therefore the original sequence -> 1/5.
 
  • #4
Thank you very much for your attention... Many thanks really...
It's been really usefull
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Mathman appears to be assuming that you want the limit as n goes to infinity which is NOT stated in the problem.

Since n, and therefore [itex]n^2[/itex] and [itex]5n^3[/itex] are fixed values, you can treat this as
[tex]\sum_{k= 0}^n\frac{A}{Bz+ 2k}[/tex]
for constant A and B and then set [itex]A= n^2[/itex] and [itex]B= 5n^3[/itex] after the sum.
If you let j= B+ 2k, this can be written
[tex]\sum_{B+2}^{B+ 2n}\frac{1}{j}[/tex]

If n= 1, this would be
[tex]\frac{1^2}{5(1^3)+ 2(0)}+ \frac{1^2}{5(1^3)+ 2(1)}= \frac{1}{5}+ \frac{1}{7}= \frac{12}{35}[/tex]

If n= 2, this would be
[tex]\frac{2^2}{5(2^3)+ 2(0)}+ \frac{2^2}{5(2^3)+ 2(1)}+ \frac{2^2}{5(2^3)+ 2(2)}=[/tex][tex] \frac{4}{40}+ \frac{4}{42}+ \frac{4}44}= \frac{1}{10}+ \frac{2}{21}+ \frac{1}{11}= \frac{551}{2310}[/tex]
 
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  • #6
Mathman appears to be assuming that you want the limit as n goes to infinity which is NOT stated in the problem.

Since n, and therefore [itex]n^2[/itex] and [itex]5n^3[/itex] are fixed values, you can treat this as
[tex]\sum_{k= 0}^n\frac{A}{Bz+ 2k}[/tex]
for constant A and B and then set [itex]A= n^2[/itex] and [itex]B= 5n^3[/itex] after the sum.
If you let j= B+ 2k, this can be written
[tex]\sum_{B+2}^{B+ 2n}\frac{1}{j}[/tex]

If n= 1, this would be
[tex]\frac{1^2}{5(1^3)+ 2(0)}+ \frac{1^2}{5(1^3)+ 2(1)}= \frac{1}{5}+ \frac{1}{7}= \frac{12}{35}[/tex]

If n= 2, this would be
[tex]\frac{2^2}{5(2^3)+ 2(0)}+ \frac{2^2}{5(2^3)+ 2(1)}+ \frac{2^2}{5(2^3)+ 2(2)}= \frac{4}{40}+ \frac{4}{42}+ \frac{4}44}= \frac{1}{10}+ \frac{2}{21}+ \frac{1}{11}= \frac{551}[/tex]{2310}

You seem to have some imprecisions...

Didn't you want to write
[tex]\sum_{B}^{B+ 2n}\frac{1}{j}[/tex]

and I suppose

[tex]\sum_{k= 0}^n\frac{A}{B+ 2k}[/tex]

Thank you very much for your attention

It's been quite useful to find out better the general term of this sequence..

Thanks a lot HallsofIvy
 
  • #7
mathman
Science Advisor
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496
Mathman appears to be assuming that you want the limit as n goes to infinity which is NOT stated in the problem.

Since n, and therefore [itex]n^2[/itex] and [itex]5n^3[/itex] are fixed values, you can treat this as
[tex]\sum_{k= 0}^n\frac{A}{Bz+ 2k}[/tex]
for constant A and B and then set [itex]A= n^2[/itex] and [itex]B= 5n^3[/itex] after the sum.
If you let j= B+ 2k, this can be written
[tex]\sum_{B+2}^{B+ 2n}\frac{1}{j}[/tex]

If n= 1, this would be
[tex]\frac{1^2}{5(1^3)+ 2(0)}+ \frac{1^2}{5(1^3)+ 2(1)}= \frac{1}{5}+ \frac{1}{7}= \frac{12}{35}[/tex]

If n= 2, this would be
[tex]\frac{2^2}{5(2^3)+ 2(0)}+ \frac{2^2}{5(2^3)+ 2(1)}+ \frac{2^2}{5(2^3)+ 2(2)}=[/tex][tex] \frac{4}{40}+ \frac{4}{42}+ \frac{4}44}= \frac{1}{10}+ \frac{2}{21}+ \frac{1}{11}= \frac{551}{2310}[/tex]
I didn't just assume that he was interested in the limit as n becomes infinite. He said it explicitly in the first post.
I've tried to solve the limit of this succession when n goes to inf, but it's been quite hard..
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
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964
You are right, mathman, I missed that. My apologies.
 
  • #9
Thanks a lot to both of you for your contributions...
Thanks
 

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