Decreasing and bounded below sequence

Question

Show that the sequence $a_n=\frac{n+1}{5^n}$ is decreasing and bounded below.

My work so far: $\large\frac{a_{n+1}}{a_n}=\frac{\frac{n+2}{5^{n+1}}}{\frac{n+1}{5^n}}$=$\frac{n+2}{5(n+1)}<1$ because $n+2<5n+5$ so we get $a_{n+1}<a_n$.

I tried to show that is bounded below by taking the real function $f(x)=\frac{x+1}{5^x}$ to find the minimum, but i don't think that's the way to do it.

Answer 1

First I want to tell you that you are confusing definitions. The work you did proves that this sequence converges, and is off topic for your aim.

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In order to prove that a sequence is bounded (with upper or lower bound), you need to pass through 2 steps:

1. You have to prove that the sequence is monotone (increasing or decreasing for all x)
2. And you have to prove that it doesn't tend to infinity, e.g. is minored (resp. majored if increasing).

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So let me begin the proof:

1. Consider that, $\forall n\in \mathbb{N},\quad { ({ a }_{ n }) }=\frac { n+1 }{ { 5 }^{ n } } $.

We will proceed by induction to prove that this sequence is indeed monotone (and decreasing in our case):

Let's show for $n=0$ and $n=1$;

$$\begin{cases} { a }_{ 0 }=\frac { 1 }{ 1 } =1 \\ { a }_{ 1 }=\frac { 2 }{ 5 } \end{cases}\Rightarrow { a }_{ 1 }<{ a }_{ 0 }$$

Then, assume that this assumption holds for $n>1$; and consider ${ a }_{ n+1 }$ to prove that ${ a }_{ n+1 }<{ a }_{ n }$:

Have we the following expression true?

$${ a }_{ n+1 }=\frac { n+2 }{ { 5 }^{ n+1 } } <\frac { n+1 }{ { 5 }^{ n } } = { a }_{ n }$$

As $n\ge 0$ because is a natural number, and for all $n$ we have: ${ 5 }^{ n }\neq 0$, through elementary calculations we obtain that $-\frac { 3 }{ 4 } <n$, which is always true.

Thus, we conclude that this sequence is a decreasing monotone sequence.

2. Now, we need to show if the sequence is minored or not (e.g. its limit through $n$ to infinity is NOT infinity). The question becomes to prove that:

$$\lim _{ n\rightarrow \infty }{ \frac { n+1 }{ { 5 }^{ n } } } \neq -\infty $$

For this, we will squeez this expression by an obvious case:

$$\frac { n }{ { 5 }^{ n } } <\frac { n+1 }{ { 5 }^{ n } } $$

So we have a simpler expression to study the case. And with usage of L'Hôpital's Rule for Limits, we will have:

$$\lim _{ n\rightarrow \infty }{ \frac { n }{ { 5 }^{ n } } } =\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n{ 5 }^{ n-1 } } } \quad =\quad 0$$

Thus, we conclude that your sequence tends to $0$ and possesses a lower bound.

$\blacksquare $

PS: And we note this situation as follows to avoid ambiguities concerning that the value $0$ is ever reached and, though, the minimum of the sequence (which is not the case here):

$$\inf { ({ a }_{ n }) } =0$$

Answer 2

Bounded below just means there is a real that is less than all the terms. You can just note that all the terms are positive, so it is bounded below by $0$. A lower bound does not have to be close. You could also say that it is bounded below by $-1000$

Answer 3

Assuming that $n$ is a natural number (which is not stated in the question), it is clear that $a_n$ is always positive since we have that $n+1>0$ and $5^n>0$ for all $n\geq 0$. Thus it is bounded below.

Answer 4

1. Prove that the sequence is strictly decreasing: $$ a_n>a_{n+1}\Leftrightarrow \frac{n+1}{5^n}>\frac{n+2}{5^{n+1}}\Leftrightarrow 5^{n+1}(n+1)>5^n(n+2)\Leftrightarrow 5n+5>n+2\\\Leftrightarrow 4n+3>0\ \ \ \forall\ \ n \in \mathbb{N} $$

2. Prove that the sequence is bounded below: $$ 0<n+2\Rightarrow 0<\frac{n+2}{5^{n+1}}\Rightarrow 0<a_{n+1}<a_n\ \ \ \forall\ \ n \in \mathbb{N} $$