Trigonometric Series Sum

[man111]

Sum of series [tex]\displaystyle \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)+ \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)+ \cos^{-1}\left(\frac{7}{\sqrt{50}}\right)+.......+ \cos^{-1}\left(\frac{(n^2-n+1)}{\sqrt{(n^2+1)(n^2-2n+2)}}\right)[/tex]

[nevrodermit]

I will prove that
\[ \sum_{i=1}^{n}{\arccos{\frac{i^2-i+1}{\sqrt{(i^2+1)(i^2-2i+2)}}}} = \arccos{\frac{1}{\sqrt{n^2+1}}}\]

Let $n=1$ then LHS is $\arccos \frac{1}{\sqrt{2}}$ and the RHS is $\arccos \frac{1}{\sqrt{2}}$
Let the statement is true for some $n>1$.
This means that
\[ \sum_{i=1}^{n}{\arccos{\frac{i^2-i+1}{\sqrt{(i^2+1)(i^2-2i+2)}}}} = \arccos{\frac{1}{\sqrt{n^2+1}}}\]
Summing both sides with $\arccos{\frac{(n+1)^2-(n+1)+1}{\sqrt{((n+1)^2+1)((n+1)^2-2(n+1)+2)}}}=\arccos{\frac{n^2+n+1}{\sqrt{(n^2+2n+2)(n^2+1)}}}$
and using that $\arccos{x}+\arccos{y}=\arccos{(xy-\sqrt{(1-x^2)(1-y^2)})}$ we get
\[\sum_{i=1}^{n+1}{\arccos{\frac{i^2-i+1}{\sqrt{(i^2+1)(i^2-2i+2)}}}} =\arccos{\frac{1}{\sqrt{n^2+1}}}+\arccos{\frac{(n^2+n+1)^2}{\sqrt{(n^2+2n+2)(n^2+1)}}} \]

\[
= \arccos{(\frac{n^2+n+1}{(n^2+1)\sqrt{n^2+2n+2}} - \sqrt{(1-\frac{1}{n^2+1})(\frac{1}{(n^2+2n+2)(n^2+1)})})}= \arccos{(\frac{1}{(n+1)^2+1})}
\]

\[
= \arccos{(\frac{n^2+n+1}{(n^2+1)\sqrt{n^2+2n+2}} - \sqrt{(\frac{n^2}{n^2+1})(\frac{1}{(n^2+2n+2)(n^2+1)})})}
\]

\[
= \arccos{(\frac{n^2+n+1}{(n^2+1)\sqrt{n^2+2n+2}} - \frac{n}{(n^2+1)\sqrt{n^2+2n+2}})}
\]
and this is
\[
\arccos{\frac{1}{\sqrt{(n+1)^2+1}}}
\]