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Consider a sequence $\{a_n\}$ of positive numbers satisfying the condition $a_na_{n+2} \le a_{n+1}^2, \forall n\in N$, then $\{a_n\}$ is a

a. convergent sequence if $a_1\ne2a_2$.

b. monotonically increasing sequence if $a_1\ne2a_2$.

c. convergent sequence if $a_1=2a_2$.

d. monotonically increasing sequence if $a_1=2a_2$.

From $a_na_{n+2} \le a_{n+1}^2$, we have $$\frac{a_{n+2}}{a_{n+1}} \le \frac{a_{n+1}}{a_n}$$. Now if we look at $$\frac{a_n}{a_1}=\frac{a_n}{a_{n-1}}.\frac{a_{n-1}}{a_{n-2}}....\frac{a_2}{a_1} \le \frac{a_2}{a_1}.\frac{a_2}{a_1}....\frac{a_2}{a_1} $$. Since the RHS goes on for $n-1$ times, $$\frac{a_n}{a_1}\le (\frac{a_2}{a_1})^{n-1}$$.

From (c), we definitely have $\{a_n\}$ as bounded as $$\frac{a_n}{a_1}\le (\frac{1}{2^{n-1}})$$. But that doesn't guarantee that the sequence is convergent. Or does it??

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$$\dfrac{a_n}{a_1}\le \dfrac{1}{2^{n-1}}\implies0<a_n\le\dfrac{1}{2^{n-1}}a_1\to0~\text{as}~n\to\infty\implies a_n\to0~\text{as}~n\to\infty$$

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a. Consider $a_{n} = n$;

b. Consider a constant sequence;

c. Consider your own work and get the $a_{1}$ to the other side of the equation. Do you see why it is convergent?

d. Consider $a_{n} = \frac{1}{2^{n}}$.

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