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.
On
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 ?
On
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.
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 $2k^2$.