$∏(1+a_n)$ converges but $∑a_n$ diverges

hbghlyj

Mathematical Analysis, Apostol著, 216页, Exercise 8.43
a) Let $a_n=(-1)^n/\sqrt n$ for $n=1,2,\dots$ Show that $∏(1+a_n)$ diverges but that $∑a_n$ converges
b) Let $a_{2n-1}=-1/\sqrt n,a_{2n}=1/\sqrt n+1/n$ for $n=1,2,\dots$ Show that $∏(1+a_n)$ converges but that $∑a_n$ diverges.
math.stackexchange.com/questions/796768/a-divergent-series-which-prod-1a-n-converges?rq=1

战巡

(a)、

\[\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}=-\eta(\frac{1}{2})=(2^{1-\frac{1}{2}}-1)\zeta(\frac{1}{2})=(\sqrt{2}-1)\zeta(\frac{1}{2})=-0.6049\]
这里
\[\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\]
为狄利克雷$\eta$函数，其与黎曼$\zeta$存在关系：$\zeta(s)(1-2^{s-1})=\eta(s)$

另一方面
\[(1+a_{2n})(1+a_{2n+1})=(1+\frac{1}{\sqrt{2n}})(1-\frac{1}{\sqrt{2n+1}})<(1+\frac{1}{\sqrt{2n}})(1-\frac{1}{\sqrt{2n}})=1-\frac{1}{2n}\]
注意很容易证明$\sum \ln(1+a_n)$和$\sum a_n$具有相同的敛散性
因此
\[\sum_{n=1}^\infty\ln(1-\frac{1}{2n})=\sum_{n=1}^\infty(-\frac{1}{2n})=-\infty\]
故有
\[\prod_{n=1}^\infty(1+a_n)<\prod_{n=1}^\infty(1-\frac{1}{2n})=0\]



(b)、
\[(1+a_{2n-1})(1+a_{2n})=(1-\frac{1}{\sqrt{n}})(1+\frac{1}{\sqrt{n}}+\frac{1}{n})=1-\frac{1}{n^{\frac{3}{2}}}\]
这里
\[\sum_{n=1}^\infty(-\frac{1}{n^{\frac{3}{2}}})=-\zeta(\frac{3}{2})\]
说明这玩意是收敛的，无穷乘积也是收敛的

而
\[a_{2n-1}+a_{2n}=\frac{1}{n}\]
这个求和当然是不收敛的