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I know that:

$$\sum_{k=1}^n\arctan(2k^2)=\frac{\pi n}{2}-\frac{1}{2}\arctan(\frac{2n(n+1)}{2n+1})$$

Can a similar closed form expressions be given for $\sum_{k=1}^n \arctan(k^2)$?

I was able to simplify it to:

$$\sum_{k=1}^n\arctan(k^2)=\frac{\pi n}{2}-\arctan(\frac{2n(n+1)}{2n+1})+\sum_{k=1}^n\arctan(\frac{1}{4k^6+3k^2})$$

But I can't simplify the sum on the right hand side.

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The sequence $\tan\left( \sum_{k=1}^n \arctan(k^2)\right)$ starts (for $n = 1$ to $15$) $$1,-\frac53,{\frac {11}{24}},-{\frac {395}{152}},{\frac {3405}{10027}},-{ \frac {364377}{112553}},{\frac {2575360}{8983513}},-{\frac {33971776}{ 9167031}},{\frac {708557735}{2760880887}},-{\frac {276796646435}{ 68094892613}},{\frac {3981342679869}{16780244555624}},-{\frac { 2420336558689725}{556533101345512}},{\frac {91633757568701803}{ 409593411519909037}},-{\frac {1960291278426118855}{428063977364527911} },{\frac {2775120694958607680}{12985105930095331479}}$$

I don't see any closed form. The numerators and denominators don't seem to be in the OEIS, nor does Maple's gfun come up with anything. There certainly isn't anything "similar" to what you have with $2k^2$.

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I am not sure that your last formula is correct. Apply it for n=1; the lhs is Pi/4 and the rhs is Pi/2 + ArcTan[1/7] - 1/2 ArcTan[4/3] = Pi/2 - ArcTan[27/4] / 4 which is more or less 1.24905.
I hope and wish I did not make any mistake.

Just by curiosity, how did you find the first summation ?

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One of the coefficients was wrong in the second sum I fixed it. – Ethan Jan 19 '14 at 8:37

Just as Robert Israel, I did not find anything for your second summation (if he does not find, I don't have many chances to find !). But, working your problem, I found (simple) that

ArcTan[k^2] + ArcTan[k^2 / (1 + 2 k^4)] = ArcTan[2 k^2]

I am almost sure that this will not help you much !

By the way, you did not tell me how you found the first summation.

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In some old stuff I had written that I just found, I can't really remember its about three years old. Though lab bhattacharjee basically gives a proof in his link. – Ethan Jan 19 '14 at 9:33

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