More on Sequences — Question 5

*Question 5 Let $a_n=\dfrac{n^2}{2^n}$ for $n\ge1$.

1. Use $a_{n+1}/a_n$ to determine exactly when $a_{n+1}<a_n$.

2. Identify the largest term and the first index from which the sequence is strictly decreasing.

3. Prove that $a_n\to0$.

*Question 5 – Solution

Because $a_n>0$, the inequality $a_{n+1}<a_n$ is equivalent to $a_{n+1}/a_n<1$. Compute $$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{2n^2}.$$

Thus $a_{n+1}<a_n$ exactly when $$(n+1)^2<2n^2\iff n^2-2n-1>0\iff n>1+\sqrt2.$$ For integer $n\ge1$, this is equivalent to $n\ge3$. Since $a_1=1/2<a_2=1<a_3=9/8$ and $a_{n+1}<a_n$ for every $n\ge3$, $a_3$ is the unique largest term and the sequence is strictly decreasing from index $3$ onward.

A decreasing positive sequence has a limit, but the ratio bound identifies it directly. For $n\ge7$, $$\frac{a_{n+1}}{a_n}=\frac12\left(1+\frac1n\right)^2\le\frac12\left(\frac87\right)^2=\frac{32}{49}<1.$$ Iterating this inequality gives $0<a_n\le a_7(32/49)^{n-7}$ for $n\ge7$. The right side is geometric with ratio less than $1$, so it tends to zero. The Squeeze Theorem yields $a_n\to0$.