$a_{2n+1}+a_{2n}=a_n$ 裂项求和

hbghlyj

实数$(a_n)_{n ≥ 1}$满足$$\tag1a_{2n+1}+a_{2n}=a_n$$则$$\tag2\sum_{n=1}^{N} a_n \cdot a_{2 n} \cdot a_{2
n+1}=\frac{1}{3} \cdot\left(a_1^{3}-\left(a_{N+1}^{3}+\ldots+a_{2 \cdot N+1}^{3}\right)\right)$$

Proof

由 (1) 与恒等式 $x^{2} \cdot y+x \cdot y^{2}=\frac{1}{3} \cdot\left((x+y)^{3}-x^{3}-y^{3}\right)$, 得\begin{array}{l}a_n\cdot a_{2n} \cdot a_{2n+1}\\=a_{2n}^{2} \cdot a_{2n+1}+a_{2n} \cdot a_{2n+1}^{2}\\=\frac{1}{3} \cdot((a_{2n+1}+a_{2n})^{3}-a_{2n}^{3}-a_{2n+1}^{3})\end{array}再使用 (1),$$a_n\cdot a_{2n} \cdot a_{2n+1}=\frac{1}{3} \cdot\left(a_n^{3}-a_{2n}^{3}-a_{2n+1}^{3}\right)$$对$n$求和, 得到(2).

hbghlyj

AOPS
\[\sum_{n=1}^{\infty} \ln\left(1+\frac{1}{n}\right) \ln \left(1+\frac{1}{2n} \right) \ln \left(1+\frac{1}{2n+1}\right).\]@auj 的解法:
易知$a_n=\ln\left(1+\frac1n\right)$满足(1).
由$\ln(1+x)<x\;\forall x>-1$得$a_n<\frac1n$, 所以
$$\sum_{n=N+1}^{2N+1}a_n^3<\frac{1}{(N+1)^{3}}+\cdots+\frac{1}{(2 \cdot N+1)^{3}}<\frac{1}{(N+1)^{3}}+\cdots+\frac{1}{(N+1)^{3}}=\frac{N+1}{(N+1)^{3}}\to0$$
代入(2)得\[\sum_{n=1}^{\infty} a_na_{2n}a_{2n+1}=\frac13(a_1^3-0)=\frac{(\ln2)^3}3\]