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.
Best Answer
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.
In order to prove that a sequence is bounded (with upper or lower bound), you need to pass through 2 steps:
So let me begin the proof:
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.
$$\lim _{ n\rightarrow \infty }{ \frac { n+1 }{ { 5 }^{ n } } } \neq -\infty $$
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):
$$\inf { ({ a }_{ n }) } =0$$