关于常数项级数的题单(20250316)¶
A:p 级数之美¶
A1:P 级数的敛散性¶
讨论 p 级数:
\[
\sum^{\infty}_{n=1}\frac{1}{n^{p}}
\]
的敛散性.
A2:对数天堂¶
利用上面的关于p 级数的讨论,尝试审敛下列级数:
-
\[\sum_{n=1}^{\infty} \frac{1}{3^{\ln n}}\]
-
\[\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}\]
-
\[\sum_{n=3}^{\infty} \frac{1}{(\ln n)^{\ln \ln n}}\]
A3:p 级数和极限还有一腿?¶
尝试计算:
\[
\lim_{n \to \infty} \left( \frac{1}{p^{n+1}} + \frac{1}{p^{n+2}} + \cdots + \frac{1}{p^{2n}} \right),p > 1
\]
B:比一比有益身体健康¶
B1:先比再说¶
利用对正项级数的比较审敛法,尝试对下列级数审敛:
-
\[\sum_{n=1}^{\infty} \left[ \frac{(2n-1)!!}{(2n)!!} \right]^k\]
-
\[\sum_{n=1}^{\infty} n! \left( \frac{a}{n} \right)^n ,a > 0\]
-
\[\sum_{n=1}^{\infty} \left[ e - \left( 1 + \frac{1}{n} \right)^n \right]\]
B2:兄弟姐妹¶
如下是常见的关于比较判别法的变式,尝试证明一下吧!假定\(a_{n}\)是正项级数.
- (Cauchy 根值判别法)若$$ \lim_{n \to \infty} \sqrt[n]{a_n} = c $$ 则:\(c<1\)时级数收敛,\(c>1\)时级数发散. $$
- (比值判别法)
- (D'lambert 判别法)若$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = d $$ 则:\(d<1\)时级数收敛,\(d>1\)时级数发散.
- (Raabe 判别法)若$$ \lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right) = r $$ 则:\(r>1\)时级数收敛,\(r<1\)时级数发散.
- (Bertrand 判别法)若$$ \lim_{n \to \infty} \ln n \left[ n \left( \frac{a_n}{a_{n+1}} - 1 \right) - 1 \right] = b $$ 则:\(b>1\)时级数收敛,\(b<1\)时级数发散.
- (Gauss 判别法)若$$ \frac{a_n}{a_{n+1}} = 1 + \frac{\mu}{n} + O\left(\frac{1}{n^{1+\varepsilon}}\right), \varepsilon > 0 $$ 则:\(\mu>1\)时级数收敛,\(\mu\leq 1\)时级数发散.