Calculus
∑n=0∞2(−1)n(n+2)(n+1)x2n+1\(\displaystyle\]\sum\)_{n=0}^{\(\infty\)}2(-1)^{n}(n+2)(n+1)\,x^{2n+1}
[−1,1]\(\left\[\lbrack\)-1,1\(\right\]\rbrack\)
∑n=0∞(−1)n(n+2)(n+1)x2n+1\(\displaystyle\]\sum\)_{n=0}^{\(\infty\)}(-1)^{n}(n+2)(n+1)\,x^{2n+1}
(−1,1)\(\left\)(-1,1\(\right\))
[−1,1)\(\left\]\lbrack\)-1,1\(\right\))