计算\(\;\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n(1+\sqrt[2^n]{2})}\)

elim

题：计算\(\;\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n(1+\sqrt[2^n]{2})}\)

elim

题：计算 \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n(1+\sqrt[2^n]{2})}.\)
解：在等式\(\small\,\dfrac{2}{1-x^2}=\dfrac{1}{1+x}+\dfrac{1}{1-x}\,\)中取\(\,x=\sqrt[2^n]{2}\,\)得\(\,(h_n=2^{-n})\)
\(\qquad\small\dfrac{1}{2^n(1+\sqrt[2^n]{2})}=\dfrac{1}{2^{n-1}(1-\sqrt[2^{n-1}]{2})}-\dfrac{1}{2^n(1-\sqrt[2^n]{2})}=\dfrac{h_{n-1}}{1-2^{h_{n-1}}}-\dfrac{h_n}{1-2^{h_n}}.\)
\(\therefore\quad\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n(1+\sqrt[2^n]{2})}={\small-1-}\lim_{n\to\infty}\frac{h_n}{1-2^{h_n}}={\small-1+}\lim_{h\to 0}\frac{1}{2^h\ln 2}=\dfrac{1}{\ln 2}\small-1.\)

elim

楼上最后一行倒数第二个等号用了 L'Hospital 法则.

elim

但愿这次 doletotodole 头不痛.

luyuanhong

楼上 elim 的帖子很好！已收藏。