# ISI2015-MMA-55

150 views

Let $\{a_n\}$ be a sequence of real numbers. Then $\underset{n \to \infty}{\lim} a_n$ exists if and only if

1. $\underset{n \to \infty}{\lim} a_{2n}$ and $\underset{n \to \infty}{\lim} a_{2n+2}$ exists
2. $\underset{n \to \infty}{\lim} a_{2n}$ and $\underset{n \to \infty}{\lim} a_{2n+1}$ exist
3. $\underset{n \to \infty}{\lim} a_{2n}, \underset{n \to \infty}{\lim} a_{2n+1}$ and $\underset{n \to \infty}{\lim} a_{3n}$ exist
4. none of the above
in Calculus
edited

## Related questions

1
195 views
The value of the infinite product $P=\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots \text{ is }$ $1$ $2/3$ $7/3$ none of the above
The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is $2$ $2e$ $2 \pi$ $2i$
The limit $\underset{n \to \infty}{\lim} \left( 1- \frac{1}{n^2} \right) ^n$ equals $e^{-1}$ $e^{-1/2}$ $e^{-2}$ $1$
Let $a_n= \bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1- \frac{1}{\sqrt{n+1}} \bigg), \: \: n \geq1$. Then $\underset{n \to \infty}{\lim} a_n$ equals $1$ does not exist equals $\frac{1}{\sqrt{\pi}}$ equals $0$