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

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

• Should options A., B., C. include wording to the effect that the limits mentioned exist? – Conrad Turner Jan 1 '16 at 6:00

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.