Proving a sequence is convergent by monotone convergence theorem (MCT)

Question

I am trying to prove the sequence $(\frac{n}{n+5})$ is convergent using only MCT.

This seems to be a trivial problem, but I cannot seem to start on a proof that the sequence is increasing. I assumed induction could be used, and started thus:

Base case:

Let $n = 1.$ Then $\frac{n}{n+5}= \frac{1}{6}$.

Inductive hypothesis:

Suppose $\frac{k}{k+5} \le \frac{k+1}{k+6}$ for some $k \in \Bbb{N}$.

Induction step:

Want to show $\frac{k+1}{k+6} \le \frac{k+2}{k+7}$.

I became stuck here. I think I am either missing some obvious algebraic manipulation, or I have not properly structured the induction problem itself.

Answer 1

Induction is not needed. If $n <m$ the $\frac 5 {m+5} <\frac 5 {n+5}$ so $1-\frac 5 {m+5} >1-\frac 5 {n+5}$. This is same as $\frac n {n+5} <\frac m {m+5}$. Hence the sequence is increasing. It is also bounded by $1$ hence convergent. (The limit is $1$).

Answer 2

Sure you can verify the sequence is increasing by checking that

$$u_{n+1}-u_{n} > 0,\forall n \in \Bbb{N}$$

or

$$\frac{u_{n+1}}{u_{n}} > 1,\forall n \in \Bbb{N}$$

However, since you want to prove the convergence, I would start to write:

$$\frac{n}{n+5}=1-\frac{5}{n+5}$$

Is now very easy to prove the sequence is monotonic and bounded:

$n$ is increasing $\Rightarrow n+5$ is increasing $\Rightarrow \frac{5}{n+5}$ is decreasing $\Rightarrow -\frac{5}{n+5}$ is increasing $\Rightarrow 1-\frac{5}{n+5}$ is increasing.

And: $\frac{1}{6} \le 1-\frac{5}{n+5}<1, \forall n \in \Bbb{N}$.