# Induction proof of the sum of the series $\sum _{k=1}^{n}\arctan\left(\frac{1}{2k^2}\right)\:$

Prove by induction that the sum

$$\sum_{k=1}^{n}\arctan\left(\frac{1}{2k^2}\right)$$

can be written as

$$S_n=\arctan\left(\frac{n}{n+1}\right).$$

I'm not quite sure what I should do here. I want to prove this for $$n = k+1$$ but I didn’t really get anywhere…

• The n in your first formula is not a variable, while it is one in your second formula. So there is a problem in your statement. If a proof by induction is required it is probably to prove that partial series are equal for all n, then find that you can compute a limit of one of the terms when $n\rightarrow \infty$ and deduce that the other one is equal.
– WNG
Apr 23 at 16:39
• What you actually want to prove is the partial sum $S_n:=\sum_{k=1}^n\arctan\frac{1}{2k^2}=\frac{n}{n+1}$.
– J.G.
Apr 23 at 17:48

Write the general term as $$\arctan\left(\frac{2}{4n^2}\right)$$ to be modified as $$\arctan\left(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right)$$ Then seperate the terms, using an angle addition formula, as $$\arctan(2n+1)-\arctan(2n-1)$$ combining the terms from 1 to n yields $$\arctan(2n+1)-\arctan(1)$$ pairing them up with the above formula gives $$\arctan\left(\frac{n}{n+1}\right)$$In case you need a limit , it is quite clearly $$\frac{\pi}{4}$$