Find $\sqrt{\frac{1}{2}-f(1)}+\dots+ \sqrt{\frac{1}{2}-f(99)}$ Define $f(n)=\sqrt[2]{n^4+\frac{1}{4}}-n^2.$
Find $$\sqrt{\frac{1}{2}-f(1)}+\dots+ \sqrt{\frac{1}{2}-f(99)}$$
I tried to simply $f(n).$ So rationalising, we get $$\sqrt{\frac{1}{2}-f(n)}\sqrt{\frac{1}{2}+f(n)}=\sqrt{\frac{1}{2}+n^2-\sqrt{n^4+\frac{1}{4}}}\sqrt{\frac{1}{2}+n^2+\sqrt[2]{n^4+\frac{1}{4}}}=\sqrt{(\frac{1}{2}+n^2)^2-n^4-\frac{1}{4}}=\sqrt{n^2}=n$$
So $$\sqrt{\frac{1}{2}-f(n)}=\frac{n}{\sqrt{\frac{1}{2}+f(n)}}.$$
EDIT: Using the hints given below, we have to find We have to find $$\sum_{n=1}^{99} \frac{n}{\sqrt{n^2+1/4}+1/2}=\sum_{n=1}^{99} \frac{n({\sqrt{n^2+1/4}-1/2})}{(\sqrt{n^2+1/4}+1/2)(\sqrt{n^2+1/4}-1/2)}=\sum_{n=1}^{99}\frac{n({\sqrt{n^2+1/4}-1/2})}{n^2}= \frac{\sqrt{n^2+1/4}-1/2}{n}$$
Any solutions?
 A: 
Claim:
$$\sqrt{\frac12-f(n)}=\frac12\left(\sqrt{(n+1)^2+n^2}-\sqrt{n^2+(n-1)^2}\right).$$

Proof:
Let
$$x=\sqrt{\frac12-f(n)}=\sqrt{\frac12+n^2-\sqrt{n^4+\frac14}},\quad y=\sqrt{\frac12+n^2\color{red}{+}\sqrt{n^4+\frac14}},$$
then by simple calculations,
\begin{align*}
\begin{cases}
x^2+y^2&=2n^2+1\\
xy&=n
\end{cases}
\end{align*}
so
\begin{align*}
\begin{cases}
y+x=\sqrt{x^2+y^2+2xy}&=\sqrt{2n^2+2n+1}\\
y-x=\sqrt{x^2+y^2-2xy}&=\sqrt{2n^2-2n+1}
\end{cases}
\end{align*}
The claim is proved by solving $x$.
Now the problem is reduced to a telescoping sum.
A: Hint
Let
$$g_n=\sqrt{\frac 12-f_n}\qquad \text{with} \qquad   f_n=\sqrt[2]{n^4+\frac{1}{4}}-n^2$$ and expand it as a series for "large" values of $n$
$$g_n=\frac 1{\sqrt 2}\Bigg[1-\frac{1}{8 n^2}+O\left(\frac{1}{n^4}\right)\Bigg]$$ Using generalized hamonic numbers
$$\sum_{n=1}^p g_n=\frac 1{\sqrt 2}\Bigg[p-\frac{1}{8}H_p^{(2)}+\cdots\Bigg]$$ Now, suppose that $p$ is large and use the asymptotics to obtain
$$\sum_{n=1}^p g_n=\frac 1{\sqrt 2}\Bigg[p-\frac{ \pi ^2}{48}+\frac 1{8p}+O\left(\frac{1}{p^2}\right)\Bigg]$$ which gives a relative error smaller than $0.01$% as soon as $p\geq 18$
