Find the value of n from the sum of a math series

\begin{aligned}
& \sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}} +...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}} = n-\frac{1}{n}\\
\end{aligned}

\begin{aligned}
n=?\\
\end{aligned}

\begin{aligned}
\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}&=\sqrt{1+\frac{1}{1}+\frac{1}{4}}\\
\\
&=\sqrt{\frac{4+4+1}{4}}\\
\\
&=\sqrt{\frac{9}{4}}\\
\\
\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}&=\frac{3}{2}\\
\end{aligned}

\begin{aligned}
\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}&=\sqrt{1+\frac{1}{4}+\frac{1}{9}}\\
\\
&=\sqrt{\frac{36+9+4}{36}}\\
\\
&=\sqrt{\frac{49}{36}}\\
\\
\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}&=\frac{7}{6}\\
\end{aligned}

\begin{aligned}
\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}&=\sqrt{1+\frac{1}{9}+\frac{1}{16}}\\
\\
&=\sqrt{\frac{144+16+9}{144}}\\
\\
&=\sqrt{\frac{169}{144}}\\
\\
\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}&=\frac{13}{12}\\
\end{aligned}

\begin{aligned}
\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}&=\sqrt{1+\frac{1}{9801}+\frac{1}{10000}}\\
\\
&=\sqrt{\frac{98010000+10000+9801}{98010000}}\\
\\
&=\sqrt{\frac{98029801}{98010000}}\\
\\
\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}&=\frac{9901}{9900}\\
\end{aligned}

\begin{aligned}
\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+....+\frac{9901}{9900}\\
\end{aligned}

\begin{aligned}
\frac{3}{2}+\frac{7}{6}&+\frac{13}{12}+....+\frac{9901}{9900}\\
\\
&= 1+\frac{1}{2}+1+\frac{1}{6}+1+\frac{1}{12}+....+1+\frac{1}{9900}\\
\\
&= (1+1+1+...1)+\bigg(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{9900}\bigg)\\
\\
&= 99+\bigg(1-\frac{1}{2}\bigg)+\bigg(\frac{1}{2}-\frac{1}{3}\bigg)+\bigg(\frac{1}{3}-\frac{1}{4}\bigg)+...+\bigg(\frac{1}{99}-\frac{1}{100}\bigg)\\
\\
&= 99+\bigg(1-\frac{1}{100}\bigg)\\
\\
&= 100 -\frac{1}{100}\\
\\
\end{aligned}

\begin{aligned}
100 -\frac{1}{100}=n -\frac{1}{n}\\
\end{aligned}