填充11.
(1)\[
\sum\limits_{n = 1}^\infty {\frac{1}{n}} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} + ...... + \frac{1}{{15}} + \frac{1}{{16}} + ......
\]
\[
> 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + ......
\]
\[
= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...... 所以發散
\]
(2)
\[
\sum\limits_{n = 1}^\infty {( - 1)^{n + 1} \frac{1}{n}} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \frac{1}{9} - \frac{1}{{10}} + ......
\]
\[
Since \ln (1 + x) = x - \frac{{x^2 }}{2} + \frac{{x^3 }}{3} - \frac{{x^4 }}{4} + \frac{{x^5 }}{5} - ...... + ( - 1)^n \frac{{x^{n + 1} }}{{n + 1}} + .....\forall x \in ( - 1,1]
\]
\[
Thus \sum\limits_{n = 1}^\infty {( - 1)^{n + 1} \frac{1}{n}} = \ln (1 + 1) = \ln 2 所以收斂
\]
如有錯誤請予指正 感謝
