Testing for series convergence.

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Homework Statement



[itex]\sum[/itex]([itex]\frac{2n}{2n+1}[/itex])n2

(The sum being from n=1 to ∞).


Homework Equations





The Attempt at a Solution



Used exponent properties to get ([itex]\frac{2n}{2n+1}[/itex])2n. Using the root test, the nth root of an = lim n->∞([itex]\frac{2n}{2n+1}[/itex])2 = 1. However, the root test is indeterminate if the limit = 1.

The ratio test didn't work out, and it doesn't seem to be similar to any series, such as geometric, tunnel, etc.

The integral test seems like it would be very complicated, but if someone thinks it's the right way to go, I'll give it a shot.

Would it work to say that ([itex]\frac{2n}{2n+1}[/itex])n2 < ([itex]\frac{2n}{2n+1}[/itex]), so if ([itex]\frac{2n}{2n+1}[/itex]) converges, then ([itex]\frac{2n}{2n+1}[/itex])n2 also does by squeeze convergence? Would that also work if it diverges?

Any help would be great, thanks!!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement



[itex]\sum[/itex]([itex]\frac{2n}{2n+1}[/itex])n2

(The sum being from n=1 to ∞).


Homework Equations





The Attempt at a Solution



Used exponent properties to get ([itex]\frac{2n}{2n+1}[/itex])2n. Using the root test, the nth root of an = lim n->∞([itex]\frac{2n}{2n+1}[/itex])2 = 1. However, the root test is indeterminate if the limit = 1.

The ratio test didn't work out, and it doesn't seem to be similar to any series, such as geometric, tunnel, etc.

The integral test seems like it would be very complicated, but if someone thinks it's the right way to go, I'll give it a shot.

Would it work to say that ([itex]\frac{2n}{2n+1}[/itex])n2 < ([itex]\frac{2n}{2n+1}[/itex]), so if ([itex]\frac{2n}{2n+1}[/itex]) converges, then ([itex]\frac{2n}{2n+1}[/itex])n2 also does by squeeze convergence? Would that also work if it diverges?

Any help would be great, thanks!!

Try a really simple test, the nth term test. If the limit of the nth term of your series as n->infinity is not zero, then the sum of the series can't exist.
 
Last edited:

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