Classic Trigonometric Olympiad Problem: Evaluate $(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 43^{\circ})(1+\tan 44^{\circ})(1+\tan 45^{\circ})$

In this blog post, we will continue to manipulate the number one to looking for the most efficient and effective solution.

According to Wikipedia (Number One):

One, sometimes referred to as unity, is the integer before two and after zero. One is the first non-zero number in the natural numbers as well as the first odd number in the natural numbers.

Show that [MATH]16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17[/MATH].

Show that [MATH]16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17[/MATH].

On my previous post (Show that [MATH]16<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{80}}<17[/MATH]), we used the trapezoid rule to prove the upper bound for the given inequality, i.e. [MATH]16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}[/MATH]. We're then not supposed to use the same method to prove the lower bound simply because the function $y=\dfrac{1}{\sqrt{x}}$ is concave up. No matter how we manipulated that concept, we will only end up with proving the target sum [MATH]\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}[/MATH] will be greater than some quantity, not less than. This works against to what we are looking to prove, that is, [MATH]\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17[/MATH].

Find The Sum Involving The Inverse Tangent Function

We are given to evaluate:

[math]S_n=\sum_{k=0}^n\left[\tan^{-1}\left(\frac{1}{k^2+k+1} \right) \right][/math]