For my Analysis 1 class, I need to solve the following problem:
Determine whether the sequence $ \left(\frac{n}{\sqrt{n+5}}\right)_{n\geq 0}$ converges and prove your answer.
MY ANSWER
I can easily show that the sequence diverges using L'Hospital's rule and I started to prove that the sequence diverges but I have a problem somewhere in my proof.
Let $\epsilon>0$ be arbitrary.
Take $N= \qquad \qquad$ (to choose at the end)
Suppose $n>N$
$\left| \frac{n}{\sqrt{n+5}} \right|= \frac{n}{\sqrt{n+5}}>\frac{n}{n+5}>\frac{N}{N+5}=\epsilon$
From there, let's find $N$ such that $\frac{N}{N+5}=\epsilon$ and then plug the value of $N$ in my initial statement.
$\begin{aligned}
\frac{N+5}{N}=\frac{1}{\epsilon} \, &\Rightarrow \, 1+\frac{5}{N}=\frac{1}{\epsilon} \\
&\Rightarrow \, \frac{5}{N}=\frac{1-\epsilon}{\epsilon} \\
&\Rightarrow \, N=\frac{5\epsilon}{1-\epsilon}
\end{aligned} \\[3ex]$
But this means that $N$ can be negative (when $\epsilon>1$) and I was wondering if it was a problem or not. I feel like it is but I'm not quite sure why.
Thanks in advance for your answers.
Best Answer
You cannot prove it that way. You proved that you always have$$\frac n{\sqrt{n+5}}>\frac n{n+5},$$but there is no way that you can deduce divergence from this. For instance, if $a_n=\frac n{n+5}+\frac 1n$, then you also have$$a_n>\frac n{n+5},$$but $(a_n)_{n\in\Bbb N}$ actually converges.
Consider the inequality $\frac n{\sqrt{n+5}}>\frac{\sqrt n}2$. When does it hold? We have\begin{align}\frac n{\sqrt{n+5}}>\frac{\sqrt n}2&\iff\sqrt{\frac n{n+5}}>\frac12\\&\iff\frac n{n+5}>\frac14,\end{align}which occurs when $n>1$. Now, for every $M>0$, if $n>4M^2$ and $n>1$, then$$\frac n{\sqrt{n+5}}>\frac{\sqrt n}2>\frac{\sqrt{4M^2}}2=M.$$And this proves that$$\lim_{n\to\infty}\frac n{\sqrt{n+5}}=\infty.$$