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Use the appropriate test to decide if they converge

  • Thread starter Natasha1
  • Start date
  • #1
467
7
Use the appropriate test to decide whether the following serie converges or not:

[tex]\sum \limit_{n=1} ^{\infty} \frac{3n^2 - 2n +1}{2n^2 + 5}[/tex]
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
859
1
Try the n-th term test for divergence.
 
  • #3
467
7
LeonhardEuler said:
Try the n-th term test for divergence.
What about these ones:

b) [tex]\sum \limit_{n=1} ^{\infty} \frac{3n^2 - 2n + 1}{2n^4 + 5}[/tex]

c) [tex]\sum \limit_{n=1} ^{\infty} \frac{n^3 4^n}{3(n!)}[/tex]

d) [tex]\sum \limit_{n=1} ^{\infty} \frac{2 + 3 sin~n}{5n^2 + 2}[/tex]
 
  • #4
LeonhardEuler
Gold Member
859
1
When you look at b, you should see that for large n it behaves like [itex]\frac{1}{2n^2}[/itex]. What test will let you use that fact?

c) is pretty straght foward. When you see everything in with powers and factorials you should think of the comparison test or the root test.

d)Similar to b). The sin basically has no effect because the series that this looks like converges absolutely.
 
  • #5
467
7
LeonhardEuler said:
When you look at b, you should see that for large n it behaves like [itex]\frac{1}{2n^2}[/itex]. What test will let you use that fact?

c) is pretty straght foward. When you see everything in with powers and factorials you should think of the comparison test or the root test.

d)Similar to b). The sin basically has no effect because the series that this looks like converges absolutely.
For a) can I just say this series does not converge as a_n does not tend to zero as n --> infinity but to 2/3 instead.

I have never heard of this n-th term test. Could someone quickly do it for a) please, just so I can understand it. Much appreciated :-)
 
  • #6
LeonhardEuler
Gold Member
859
1
The n-th term test can be used exactly like you said. If the terms in the sequence do not approach zero then the series can not converge. That is all you need to say.
 

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