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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…

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    $\begingroup$ 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. $\endgroup$
    – WNG
    Apr 23 at 16:39
  • $\begingroup$ What you actually want to prove is the partial sum $S_n:=\sum_{k=1}^n\arctan\frac{1}{2k^2}=\frac{n}{n+1}$. $\endgroup$
    – J.G.
    Apr 23 at 17:48

1 Answer 1

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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 $\frac{\pi}{4}$

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  • $\begingroup$ For the induction aspect, you can break down the nth term into the series sum and use that for the n+1 th case $\endgroup$
    – Sung Jin-Woo
    Apr 23 at 19:18

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