Applying comparison theorem to limits when inequality is known I somehow confuse myself whenever I apply the comparison lemma.  I know there is the comparison lemma that says the following:

Let the sequence $\{a_{n}\}$ converge to the number $a$. Then the sequence $\{b_{n}\}$ converges to the number $b$ if there is a non-negative number $C$ and an index $N$ such that $$\lvert b_{n} - b \rvert \leq C \lvert a_{n}-a\rvert$$

Somehow i accidentally apply this wrong. For example let's say I'm applying it to the sequence $a_n$, to say that it converges to $0$,
$$a_n=\frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}$$
and
$$b_n=\frac{1}{\sqrt{1+\frac{1}{n}}+1}$$
Obviously, since $n\ge1$, then I can say
$$\frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}<\frac{1}{\sqrt{1+\frac{1}{n}}+1}$$
Subtracting $1/2$ from both sides, I can see that
$$\frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}-\frac{1}{2}<\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2}$$
Since I know that $$\frac{1}{\sqrt{1+\frac{1}{n}}+1} \longrightarrow \frac{1}{2}$$  then the comparison lemma seems to be telling me that $a_n$ converges to $1/2$, when in fact I know that it converges to $0$.  Where am I going wrong in this logic?
 A: The problem is that $$\sqrt{1+\frac{1}{n}} + 1 > 2$$ for all $n\in\mathbb N$, and therefore
$$\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2} < 0$$
When you compare the absolute values of both sides of your last inequality, the "$<$" becomes "$>$", and the theorem does not apply.
Namely, because the values inside the absolute value sign are negative, we have that
$$\left| \frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)} - \frac{1}{2} \right| = \frac{1}{2} - \frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}$$
$$\left| \frac{1}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{2} \right| = \frac{1}{2} - \frac{1}{\sqrt{1+\frac{1}{n}}+1}$$
But, starting from the inequality (that does indeed hold):
$$\frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}-\frac{1}{2}<\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2}$$
and multiplying both sides by -1, and remembering that we have to flip the inequality sign, we arrive at
$$\left| \frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}-\frac{1}{2} \right| > \left|\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2} \right|$$
Therefore, the sequences $a_n$ and $b_n$ do not satisfy the conditions of the theorem.
