Prove π4+16>arctan(65).
Let's pick up where we left off...we needed to complete the proof using the readily available formula that says
arctan(65)=π4+110−1100+11500−1125000+1750000−18750000+⋯
such that we get
π4+16>arctan(65)
Okay, it's very tempting for students to immediately group the readily available formula in such a way that it is now beautifully grouped:
arctan(65)=π4+110−1100+11500−1125000+1750000−18750000+⋯
arctan(65)=π4+(110−1100)+(11500−1125000)+(1750000−18750000)+⋯
Up to this juncture, what the above can tell us is that each group term is greater than zero, i.e. arctan(65) is the result of the sum of π4 plus a series of other positive small values, in other words, that also means arctan(65)>π4
Wait a minute! This isn't particularly useful to prove the question in hand!
We therefore need to devise a plan. If we can prove
\displaystyle \small \color{yellow}\bbox[5px,purple]{\text{some value}}\color{black}\gt \arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots
and \displaystyle \color{black}\dfrac{\pi}{4}+\dfrac{1}{6}\gt \color{yellow}\bbox[5px,purple]{\text{that value}}
then we have \displaystyle \color{black}\dfrac{\pi}{4}+\dfrac{1}{6}\gt \color{yellow}\bbox[5px,purple]{\text{that value}}\color{black} \gt \arctan\left({\dfrac{6}{5}}\right) and the result is then followed.
That is to say, we need to rearrange and regrouping what we have on the RHS of the equality where
\arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots
in such a special way so that \arctan\left({\dfrac{6}{5}}\right) is less than some value...
Think of it if we regroup the term so that each pair of the group terms are all negatives:
\displaystyle \color{black}\arctan\left({\dfrac{6}{5}}\right)=\color{yellow}\bbox[5px,purple]{\dfrac{\pi}{4}+\dfrac{1}{10}}\color{black}-\left(\dfrac{1}{100}-\dfrac{1}{1500}\right)-\left(\dfrac{1}{125000}-\dfrac{1}{750000}\right)-\cdots
Yes! At this point, we can safely say that
\displaystyle \color{yellow}\bbox[5px,purple]{\dfrac{\pi}{4}+\dfrac{1}{10}}\color{black}\gt \arctan\left({\dfrac{6}{5}}\right)
And note that
\displaystyle \color{black}\dfrac{\pi}{4}+\dfrac{1}{6}\gt \color{yellow}\bbox[5px,purple]{\dfrac{\pi}{4}+\dfrac{1}{10}}
We can now put them together that yields:
\displaystyle \color{black}\dfrac{\pi}{4}+\dfrac{1}{6}\gt \color{yellow}\bbox[5px,purple]{\dfrac{\pi}{4}+\dfrac{1}{10}}\color{black}\gt \arctan\left({\dfrac{6}{5}}\right)
and therefore, evidently we have proved that:
\dfrac{\pi}{4}+\dfrac{1}{6}\gt \arctan\left({\dfrac{6}{5}}\right).
This isn't the sole method that one could use to attack the problem, we could also approach it like follows:
I know I know, many students would add \dfrac{1}{6} and at the same time subtract the fraction \dfrac{1}{6} to the right side of the equality and wind up with
\small \arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{6}-\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots
\small \arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{6}-\left(\dfrac{1}{6}-\dfrac{1}{10}\right)-\left(\dfrac{1}{100}-\dfrac{1}{1500}\right)-\left(\dfrac{1}{125000}-\dfrac{1}{750000}\right)-\cdots
Note that each parenthesis pair that consists of the subtraction of a smaller value from a bigger value will definitely gives a positive value, that means:
\small \arctan\left({\dfrac{6}{5}}\right)+\left(\dfrac{1}{6}-\dfrac{1}{10}\right)+\left(\dfrac{1}{100}-\dfrac{1}{1500}\right)+\left(\dfrac{1}{125000}-\dfrac{1}{750000}\right)+\cdots=\dfrac{\pi}{4}+\dfrac{1}{6}
which yields:
\dfrac{\pi}{4}+\dfrac{1}{6}\gt \arctan\left({\dfrac{6}{5}}\right) and we are hence done.
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