CONTEXT: Uni question made up by lecturer
How would use the Taylor polynomial to degree 5 to approximate $4\arctan(\frac{1}{5})-\arctan(\frac{1}{239})$
Recall Machin's formula: $\frac{\pi}{4}=4\arctan(\frac{1}{5})-\arctan(\frac{1}{239})$ .
I tried to do this using a difference of $\arctan x=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1}$, where I subbed $x=\frac{1}{5}$ and $x=\frac{1}{239}$ and multiplied the first sigma by $4$, but couldn't work out how to simplify two infinite sums into one.
I was left with: $$4\arctan(\frac{1}{5})-\arctan(\frac{1}{239})=\sum_{n=0}^\infty \frac{4(-1)^n}{5^{2n+1}(2n+1)}-\sum_{n=0}^\infty \frac{(-1)^n}{239^{2n+1}(2n+1)}$$
If you really want a single sum, reduce to same denominator to get $$\sum_{n=0}^\infty \frac{(-1)^n \left(4\times 239^{2 n+1}-5^{2 n+1}\right)}{(2 n+1)\,1195^{2n+1}}$$