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Let $a_n$ be a sequence of real numbers. Then $\displaystyle\lim_{n \to \infty}a_n$ exists iff

A. $\displaystyle\lim_{n \to \infty}a_{2n}$ and $\displaystyle\lim_{n \to \infty}a_{2n+2}$ exists

B.$\displaystyle\lim_{n \to \infty}a_{2n+1}$ and $\displaystyle\lim_{n \to \infty}a_{2n}$ exists

C.$\displaystyle\lim_{n \to \infty}a_{2n}$ , $\displaystyle\lim_{n \to \infty}a_{2n+1}$, and $\displaystyle\lim_{n \to \infty}a_{3n}$ exists

D. None of Above

I need to think of sequence such that conditions are options are violated but limit exists to remove options. But i am not coming up with such

Thanks for help

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    $\begingroup$ Should options A., B., C. include wording to the effect that the limits mentioned exist? $\endgroup$ – Conrad Turner Jan 1 '16 at 6:00
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Hint: For A and B, consider $a_n = (-1)^n$.

Then, $\displaystyle\lim_{n \to \infty}a_{2n} = 1$, $\displaystyle\lim_{n \to \infty}a_{2n+1} = -1$, and $\displaystyle\lim_{n \to \infty}a_{2n+2} = 1$, but $\displaystyle\lim_{n \to \infty}a_{n}$ doesn't exist.

For C, look at this question.

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  • $\begingroup$ So Answer is C ? $\endgroup$ – Sophie Clad Jan 1 '16 at 6:07
  • $\begingroup$ ^Yes, that is correct. $\endgroup$ – JimmyK4542 Jan 1 '16 at 6:15

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