Determining if the sequence convergers or diverges(II)

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SUMMARY

The sequence defined by the expression $$\frac{n^2}{2n - 1} - \frac{n^2}{2n + 1}$$ converges to $$\frac{1}{2}$$ as n approaches infinity. The correct approach to determine convergence is to simplify the expression and take the limit, rather than attempting to analyze it as a telescoping series. The limit calculation shows that the sequence approaches $$\frac{2}{4 - 0}$$, confirming convergence to $$\frac{1}{2}$$.

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shamieh
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I'm confused on how they are getting their result...

Determine if the sequence converges or diverges, if it converges, find the limit...

$$\frac{n^2}{2n - 1} - \frac{n^2}{2n + 1}$$

So I started plugging in from 1 because it looks like they want me to do something with a telescoping series and I got:

(1 - 1/3) + (4/3 - 4/5) + (9/5 - 9/7) + ... (Which really got me no where?)

Then I tried to just take the limit as n -> infinity using l'opitals and ended up with $$n - n$$ which got me no where as well...So I'm really confused on how they know that it converges to $$\frac{1}{2}$$
 
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shamieh said:
I'm confused on how they are getting their result...

Determine if the sequence converges or diverges, if it converges, find the limit...

$$\frac{n^2}{2n - 1} - \frac{n^2}{2n + 1}$$

So I started plugging in from 1 because it looks like they want me to do something with a telescoping series and I got:

(1 - 1/3) + (4/3 - 4/5) + (9/5 - 9/7) + ... (Which really got me no where?)

Then I tried to just take the limit as n -> infinity using l'opitals and ended up with $$n - n$$ which got me no where as well...So I'm really confused on how they know that it converges to $$\frac{1}{2}$$

What you've been given is NOT a series, so why would you try to see if this non-existent series is telescopic?

The right thing to do with this sequence is to take the limit as n approaches infinity.

$\displaystyle \begin{align*} \frac{n^2}{2n -1} - \frac{n^2}{2n + 1} &= \frac{n^2 \left( 2n + 1 \right) - n^2 \left( 2n - 1 \right) }{ \left( 2n -1 \right) \left( 2n + 1 \right) } \\ &= \frac{2n^3 + n^2 - 2n^3 + n^2}{4n^2 - 1} \\ &= \frac{2n^2}{4n^2 - 1} \\ &= \frac{2}{4 - \frac{1}{n^2} } \\ &\to \frac{2}{4 - 0} \textrm{ as } n \to \infty \\ &= \frac{1}{2} \end{align*}$

So the sequence converges to 1/2.
 
Last edited:
Prove It said:
What you've been given is NOT a series, so why would you try to see if this non-existent series is telescopic?So the series converges to 1/2.

Yes you are correct I am an idiot. Idk why I tried that. I'm assuming you mean the sequence converges though right? Just a typo? Again, thank you for the explanation. Helped me so much!
 
Yes that was a typo, I fixed it now :)
 

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