Let $ a \in \mathbb{N}^\ast $, prove that $ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $ I have the following problem to solve. It's about convergent sequences.
Let $ a \in \mathbb{N}^\ast $, prove that: 
$$ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $$
To solve it, I first solve:
$$ 0 < \sqrt{a^{2}+1} - a - \frac{1}{2a} + \frac{1}{2a^3} $$
To do so, I compute the result for $ a = 1 $ which is $ \sqrt{2} - \frac{11}{8} > 0$. Then I was thinking of studying the variation of the sequence with it's derivative but it makes weird equations:
$$ U_a = \sqrt{a^{2}+1} - a - \frac{1}{2a} + \frac{1}{2a^3} $$
$$ U_a' = \frac{a}{\sqrt{a^2 + 1}} - 1 + \frac{1}{2a^2} - \frac{3}{2a^4} $$ 
$$ U_a' = 0 \Leftrightarrow \frac{2a^5 \sqrt{a^2+1} + 2a^4}{7a^2+9} - 1 = 0 $$ 
I'm lost from here, someone can help me ?
If it helps, we are studying limited expansions.
Thank you.
 A: Note that $$\sqrt{a^2+1}<\sqrt{a^2+1+\frac{1}{4a^2}}=\sqrt{\left(a+\frac1{2a}\right)^2}=a+\frac1{2a}.$$
Thus
$$\sqrt{a^2+1}-a<\frac{1}{2a}.$$
On the other hand
$$\sqrt{a^2+1}+a<2a+\frac{1}{2a}.$$
Therefore
$$\sqrt{a^2+1}-a=\frac{1}{\sqrt{a^2+1}+a}>\frac{1}{2a+\frac{1}{2a}}.$$
However
$$\frac{1}{2a+\frac{1}{2a}}=\frac{1}{2a\left(1+\frac{1}{4a^2}\right)}>\frac{1-\frac{1}{(4a^2)^2}}{2a\left(1+\frac{1}{4a^2}\right)}=\frac{1}{2a}\left(1-\frac{1}{4a^2}\right).$$
Hence we actually have a stronger inequality
$$\sqrt{a^2+1}-a>\frac{1}{2a}-\frac{1}{8a^3}>\frac{1}{2a}-\frac{1}{2a^3}.$$
A: The Maclaurin series of $\sqrt{x^2+1}$:$$\sqrt{x^2+1}=1+\dfrac{1}{2}x^2-\dfrac{1}{8}x^4+\dfrac{1}{16}x^5-\cdots$$ Then, find $\dfrac{\sqrt{x^2+1}-1}{x}$,$$\dfrac{\sqrt{x^2+1}-1}{x}=\dfrac{1}{2}x-\dfrac{1}{8}x^3+\dfrac{1}{16}x^5-\cdots$$ Substitute $x=\dfrac{1}{a}$:$$\dfrac{\sqrt{x^2+1}-1}{x}=\dfrac{\sqrt{\frac{1}{a^2}+1}-1}{\frac{1}{a}}=\sqrt{a^2+1}-a=\dfrac{1}{2a}-\dfrac{1}{8a^3}+\dfrac{1}{16a^5}-\cdots$$
Then, obviously, $$\dfrac{1}{2a}-\dfrac{1}{2a^3}<\dfrac{1}{2a}-\dfrac{1}{8a^3}<\sqrt{a^2+1}-a<\dfrac{1}{2a}$$
