(a)
매우 큰 k k k k ≪ k 2 k\ll k^2 k ≪ k 2 a n = k 3 k 2 − k ≈ k 3 k 2 = 1 k 5 / 3 a_n=\frac{\sqrt[3]{k}}{k^2-k}\approx \frac{\sqrt[3]{k}}{k^2}
=\frac{1}{k^{5/3}} a n = k 2 − k 3 k ≈ k 2 3 k = k 5/3 1 b k = 1 k 5 / 3 b_k=\frac{1}{k^{5/3}} b k = k 5/3 1
lim k → ∞ a k b k = lim k → ∞ k 3 / ( k 2 − k ) 1 / k 5 / 3 = lim k → ∞ k 2 k 2 − k = lim k → ∞ 1 1 − 1 / k = 1 \begin{aligned}
\lim_{k\rightarrow\infty}\frac{a_k}{b_k}
=\lim_{k\rightarrow\infty}\frac{\sqrt[3]{k}/(k^2-k)}{1/k^{5/3}}
=\lim_{k\rightarrow\infty}\frac{k^2}{k^2-k}
=\lim_{k\rightarrow\infty}\frac{1}{1-1/k}
=1
\end{aligned} k → ∞ lim b k a k = k → ∞ lim 1/ k 5/3 3 k / ( k 2 − k ) = k → ∞ lim k 2 − k k 2 = k → ∞ lim 1 − 1/ k 1 = 1
극한이 존재하고 0이 아니며, ∑ k = 1 ∞ 1 k 5 / 3 \sum\limits_{k=1}^\infty\frac{1}{k^{5/3}} k = 1 ∑ ∞ k 5/3 1 p p p p = 5 3 > 1 p=\frac{5}{3}>1 p = 3 5 > 1
(b)
이 급수의 k t h k^{\rm th} k th a k = k 10 10 k ( k ! ) 2 ( 2 k ) ! a_k= \frac{k^{10}10^k(k!)^2}{(2k)!} a k = ( 2 k )! k 10 1 0 k ( k ! ) 2
a k + 1 a k = ( k + 1 ) 10 10 k + 1 ( ( k + 1 ) ! ) 2 ( 2 k + 2 ) ! ⋅ ( 2 k ) ! k 10 10 k ( k ! ) 2 = 10 ( k + 1 k ) 10 ( k + 1 ) 2 ( 2 k + 2 ) ( 2 k + 1 ) = 10 ( 1 + 1 k ) 10 ( 1 + 1 / k ) 2 ( 2 + 2 / k ) ( 2 + 1 / k ) \begin{aligned}
\frac{a_{k+1}}{a_k}
&=\frac{(k+1)^{10}10^{k+1}((k+1)!)^2}{(2k+2)!}\cdot\frac{(2k)!}{k^{10}10^k(k!)^2}
=10\Big(\frac{k+1}{k}\Big)^{10}\frac{(k+1)^2}{(2k+2)(2k+1)}\\
&=10\Big(1+\frac{1}{k}\Big)^{10}\frac{(1+1/k)^2}{(2+2/k)(2+1/k)}
\end{aligned} a k a k + 1 = ( 2 k + 2 )! ( k + 1 ) 10 1 0 k + 1 (( k + 1 )! ) 2 ⋅ k 10 1 0 k ( k ! ) 2 ( 2 k )! = 10 ( k k + 1 ) 10 ( 2 k + 2 ) ( 2 k + 1 ) ( k + 1 ) 2 = 10 ( 1 + k 1 ) 10 ( 2 + 2/ k ) ( 2 + 1/ k ) ( 1 + 1/ k ) 2
k k k ∞ \infty ∞ 10 × 1 × 1 2 × 2 > 1 10\times 1\times\frac{1}{2\times 2}>1 10 × 1 × 2 × 2 1 > 1
(c)
적분판정법을 사용한다. 이 급수의 k t h k^{\rm th} k th a k = 1 k ( log k ) ( log log k ) = f ( k ) a_k=\frac{1}{k(\log k) (\log\log k)}=f(k) a k = k ( l o g k ) ( l o g l o g k ) 1 = f ( k ) f ( x ) = 1 x ( log x ) ( log log x ) f(x)=\frac{1}{x(\log x) (\log\log x)} f ( x ) = x ( l o g x ) ( l o g l o g x ) 1 x ≥ 3 x\ge 3 x ≥ 3
∫ 3 ∞ f ( x ) d x = ∫ 3 ∞ d x x ( log x ) ( log log x ) = lim R → ∞ ∫ 3 R d x x ( log x ) ( log log x ) = lim R → ∞ ∫ log 3 log R d y y log y 여기서 y = log x , d y = d x x = lim R → ∞ ∫ log log 3 log log R d t t 여기서 t = log y , d t = d y y = lim R → ∞ [ log t ] log log 3 log log R = ∞ \begin{aligned}
\int_3^\infty f(x)\ \mathrm{d}x
&= \int_3^\infty \frac{\mathrm{d}x}{x(\log x) (\log\log x)}
=\lim_{R\rightarrow\infty} \int_3^R \frac{\mathrm{d}x}{x(\log x) (\log\log x)}
\\[0.1in]
&=\lim_{R\rightarrow\infty} \int_{\log 3}^{\log R} \frac{\mathrm{d}y}{y\log y}
&\qquad\text{여기서 }y=\log x,\ \mathrm{d}y=\frac{\mathrm{d}x}{x} \\[0.1in]
&= \lim_{R\rightarrow\infty}\int_{\log \log 3}^{\log \log R} \frac{\mathrm{d}t}{t}
& \qquad\text{여기서 }t=\log y,\ \mathrm{d}t=\frac{\mathrm{d}y}{y} \\[0.1in]
&=\lim_{R \to \infty}\Big[\log t\Big]_{\log \log 3}^{\log\log R}=\infty
\end{aligned} ∫ 3 ∞ f ( x ) d x = ∫ 3 ∞ x ( log x ) ( log log x ) d x = R → ∞ lim ∫ 3 R x ( log x ) ( log log x ) d x = R → ∞ lim ∫ l o g 3 l o g R y log y d y = R → ∞ lim ∫ l o g l o g 3 l o g l o g R t d t = R → ∞ lim [ log t ] l o g l o g 3 l o g l o g R = ∞ 여기서 y = log x , d y = x d x 여기서 t = log y , d t = y d y
적분이 발산하므로 적분판정법에 의해 이 급수도 발산한다.