What is general approach to determine sequence is unbounded?

Question

Prove that the following sequences are unbounded: $$x_{n} = \frac{2^n}{n^2}$$ $$x_{n} = \frac{3^n - 2^n}{2^n + 1}$$ How can I do this strictly and elegant? Explanation will be appreciated.

Answer 1

Let's consider any $a>1$ $$\frac{n}{a^n}=\frac{n}{[1+(a-1)]^n}<\frac{2n}{n(n-1)(a-1)^2} \to 0$$ now for any $k \in \mathbb{N}$ $$\frac{n^k}{a^n}=\left( \frac{n}{\sqrt[k]{a^n}}\right)^k=\left(\frac{n}{(\sqrt[k]{a})^n} \right)^k = \left(\frac{n}{b^n} \right)^k$$ where $b=\sqrt[k]{a}>1.$

In first your example we will have $\lim\limits_{n \to \infty}\frac{n^2}{2^n}=0 \Rightarrow \lim\limits_{n \to \infty}\frac{2^n}{n^2}=\infty$, so it's unbounded.

In second example $\lim\limits_{n \to \infty}\frac{2^n + 1}{3^n - 2^n} = \lim\limits_{n \to \infty}\frac{1}{3^n}\frac{\frac{2^n}{3^n} + \frac{1}{3^n}}{1 - \frac{2^n}{3^n}} = 0\Rightarrow\lim\limits_{n \to \infty}\frac{3^n - 2^n}{2^n + 1}=\infty$, so, again, it's unbounded.

Answer 2

$$n>4\implies\frac{2^{n+1}}{(n+1)^2}>\frac{2.2^n}{\tfrac{3}{2}n^2}>(\tfrac{4}{3})^na_4 \textrm{ (by induction)}$$

$$n>1\implies\frac{3^n-2^n}{2^n+1}>\frac{\tfrac{1}{2}3^n}{2.2^n}=\tfrac{1}{4}(1.5)^n$$

The general method is to try to see which terms dominate and then 'reduce' the other terms to them, thus resulting in one fractional power.

Answer 3

Well one way in cases like these which are growing geometrically is to use the ratio of successive terms.

For the first, for example $$\frac {x_{n+1}}{x_n}=\frac {2^{n+1}}{(n+1)^2}\frac {n^2}{2^n}=\frac 2{\left(1+\frac 1n\right)^2}$$

Now if you can show that this is eventually greater than or equal to say $\frac 98\gt 1$ ($n=3$) then $x_n$ grows from some point faster than a constant times a power of $\frac 98$.

Answer 4

$$\frac{\dfrac{2^{n+1}}{(n+1)^2}}{\dfrac{2^n}{n^2}}=2\frac{n^2}{(n+1)^2}\ge\frac98$$ for $n\ge3$ and the sequence grows faster than a diverging geometric progression.

$$\dfrac{2^n}{n^2}\ge\left(\frac 98\right)^{n-4}.$$

---

$$\frac{3^n-2^n}{2^n-1}=\frac{3^n-1}{2^n-1}-1>\frac{3^n-1}{2^n}-1>\frac{3^n}{2^n}-2$$ which is unbounded.

---

Note that if $a>b$,

$$a^n\pm \lambda b^n\sim a^n.$$

Answer 5

For the first sequence, the standard way uses the ratio and Bernoulli's inequality: $$\frac{x_{n+1}}{x_n}=\frac{2^{n+1}}{2^n}\,\frac{n^2}{(n+1)^2}=2\Bigl(1-\frac 1{n+1}\Bigr)^2>2\Bigl(1-\frac 2{n+1}\Bigr)=2-\frac 4{n+1},$$ so that if, say, $n\ge 9$, $\:\frac{x_{n+1}}{x_n} > 2-0.4>\frac32$.

Therefore, an induction for nursery schools shows that $$x_n>\Bigl(\frac32\Bigr)^{\!n-9}\!\! x_9,\quad\text{which tends to }\infty.$$

For the second sequence, I'd use a high-school identity: \begin{align} \frac{3^n-2^n}{2^n+1}&=\frac{(3-2)(3^{n-1}+2\cdot3^{n-2}+\dots+2^{n-2}\cdot 3+2^{n-1}}{2^n+1} \\ &>\frac{n\cdot 2^{n-1}}{2\cdot2^n}=\frac n4. \end{align}