- #1

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well as n->inf, cos goes to 1 right? so shouldnt the limit of this sequence be 0?

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- #1

- 140

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well as n->inf, cos goes to 1 right? so shouldnt the limit of this sequence be 0?

- #2

saltydog

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ProBasket said:

well as n->inf, cos goes to 1 right? so shouldnt the limit of this sequence be 0?

Mathematica returns:

[tex]\mathop \lim\limits_{n\to \infty}n^2[1-Cos(\frac{a}{n})]=\frac{a^2}{2}[/tex]

I'd like to know how too.

- #3

shmoe

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[tex]a_n=\frac{1-\cos{\frac{5.2}{n}}}{\frac{1}{n^2}}[/tex]

and use l'hopital, or you could use the taylor series for cos.

- #4

saltydog

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shmoe said:

[tex]a_n=\frac{1-\cos{\frac{5.2}{n}}}{\frac{1}{n^2}}[/tex]

and use l'hopital, or you could use the taylor series for cos.

Right, just twice. Thanks.

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