Materi : Deret tak terhingga
Ujilah tiap² deret untuk konvergensi mutlak atau bersyarat :
[tex] \frac{1}{3 \times {2}^{2} } - \frac{1}{4 \times {3}^{2} } + \frac{1}{5 \times {4}^{2} } - \frac{1}{6 \times {5}^{2} } + ....[/tex]
Percayalah suatu saat orang yang kita sukai atau percayai akan mengkhianati kita:)
Jawaban:
[tex]\large\text{$\begin{aligned}&\frac{1}{3\times2^2}-\frac{1}{4\times3^2}+\frac{1}{5\times4^2}-\frac{1}{6\times5^2}+{\dots}\\&{=\ }\sum_{n=1}^{\infty}\left[(-1)^{n+1}\frac{1}{(n+2)(n+1)^2}\right]\quad.....(i)\\\\\end{aligned}$}[/tex]
Deret tersebut adalah deret ganti tanda (DGT) atau alternating series.
Deret ganti tanda, dengan bentuk umum [tex]\large\text{$\sum_{n=1}^{\infty}{\left[(-1)^{n+1}\cdot a_n\right]}$}[/tex] dikatakan konvergen jika:
[tex]\large\text{$\begin{array}{ll}1.&a_n\ \textsf{monoton turun}\\2.&\lim\limits_{n\to\infty}{a_n}=0\end{array}$}[/tex]
Dari (i), kita peroleh:
[tex]\large\text{$\begin{aligned}&\begin{cases}U_n=(-1)^{n+1}\dfrac{1}{(n+2)(n+1)^2}\\\\\left|U_n\right|=a_n=\dfrac{1}{(n+2)(n+1)^2}\end{cases}\end{aligned}$}[/tex]
sehingga [tex]a_n=\dfrac{1}{(n+2)(n+1)^2}[/tex] dan [tex]a_{n+1}=\dfrac{1}{(n+3)(n+2)^2}[/tex]
[tex]\large\text{$\begin{aligned}\frac{a_n}{a_{n+1}}&=\frac{\ \dfrac{1}{(n+2)(n+1)^2}\ }{\dfrac{1}{(n+3)(n+2)^2}}\\\\&=\frac{(n+3)(n+2)^2}{(n+2)(n+1)^2}\\\\&=\frac{(n+3)(n+2)}{(n+1)^2}\\\\&=\frac{(n+1)^2+3n+5}{(n+1)^2}\\\\&=1+\frac{3n+5}{(n+1)^2}\\\\\frac{a_n}{a_{n+1}}&>1\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}a_{n+1} < a_n\implies\textsf{$a_n$ turun.}\end{aligned}$}[/tex]
Kemudian,
[tex]\large\text{$\begin{aligned}\lim_{n\to\infty}{a_n}&=\lim_{n\to\infty}{\frac{1}{(n+2)(n+1)^2}}\\&=\frac{\lim\limits_{n\to\infty}1}{\lim\limits_{n\to\infty}(n+2)(n+1)^2}\\&=\frac{\lim\limits_{n\to\infty}1}{\lim\limits_{n\to\infty}(n+2)\cdot\lim\limits_{n\to\infty}(n+1)^2}\\\lim_{n\to\infty}{a_n}&=0\end{aligned}$}[/tex]
Karena kedua syarat terpenuhi, maka deret tersebut konvergen.
Konvergen Mutlak dan Bersyarat
[tex]\large\text{$\begin{aligned}&\textsf{Misalkan $\sum_{n=1}^{\infty}U_n\,$ dan $\forall\,{U_n}\ne0\,$.}\\&\textsf{(i)\:\:Jika $\sum_{n=1}^{\infty}\left|U_n\right|$ konvergen, maka:}\\&{\quad\:\ }\textsf{$\sum_{n=1}^{\infty}U_n$ \underline{konvergen mutlak}.}\\&\textsf{(ii) Jika $\sum_{n=1}^{\infty}\left|U_n\right|$ divergen, namun}\\&{\quad\:\ }\textsf{$\sum_{n=1}^{\infty}U_n$ konvergen, maka:}\\&{\quad\:\ }\textsf{$\sum_{n=1}^{\infty}U_n$ \underline{konvergen bersyarat}.}\\\end{aligned}$}[/tex]
Karena suku-suku pada [tex]\begin{aligned}\sum_{n=1}^{\infty}\left|U_n\right|\end{aligned}[/tex] semuanya positif, maka kita lakukan Uji Deret Positif.
Kita gunakan Uji Banding Limit, karena perpangkatan pada pembagi tetap.
[tex]\large\text{$\begin{aligned}&L=\lim_{n\to\infty}\frac{a_n}{b_n}\\&\textsf{Untuk $b_n$, dipilih: $b_n=\frac{1}{n^3}$\ .}\\\\&L=\lim_{n\to\infty}\frac{\ \dfrac{1}{(n+2)(n+1)^2}\ }{\dfrac{1}{n^3}}\\\\&{\ \;}=\lim_{n\to\infty}\frac{n^3}{(n+2)(n+1)^2}\\\\&{\ \;}=\lim_{n\to\infty}\frac{n^3}{n\left(1+\frac{2}{n}\right)\left(n\left(1+\frac{1}{n}\right)\right)^2}\\\\&{\ \;}=\lim_{n\to\infty}\frac{\cancel{n^3}}{\cancel{n^3}\left(1+\frac{2}{n}\right)\left(1+\frac{1}{n}\right)^2}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&L=\lim_{n\to\infty}\frac{1}{\left(1+\frac{2}{n}\right)\left(1+\frac{1}{n}\right)^2}\\\\&\quad\left[\ \normalsize\textsf{dengan pengecualian bentuk tak tentu} \right.\\&{\ \;}=\frac{\lim\limits_{n\to\infty}1}{\lim\limits_{n\to\infty}\left(1+\frac{2}{n}\right)\cdot\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^2}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&{\ \;}=\frac{\lim\limits_{n\to\infty}1}{\left(\lim\limits_{n\to\infty}1+\lim\limits_{n\to\infty}\frac{2}{n}\right)\cdot\left(\lim\limits_{n\to\infty}1+\lim\limits_{n\to\infty}\frac{1}{n}\right)^2}\\\\&{\ \;}=\frac{1}{(1+0)\cdot(1+0)^2}=\frac{1}{1\cdot1}\\\\&L=1\end{aligned}$}[/tex]
Nilai L = 1, artinya 0 < L < ∞.
Jika [tex]\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{n^3}\end{aligned}[/tex] konvergen, maka deret yang kita uji juga konvergen.
Uji deret-p menyatakan bahwa jika p > 1, maka deret [tex]\begin{aligned}\sum_{n=1}^{\infty}{\frac{1}{n^p}}\end{aligned}[/tex] konvergen.
Karena pada [tex]\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{n^3}\end{aligned}[/tex] nilai p adalah 3, artinya p > 1, maka deret tersebut konvergen.
Karena [tex]\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{n^3}\end{aligned}[/tex] konvergen, berdasarkan Uji Banding Limit, [tex]\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{(n+2)(n+1)^2}\end{aligned}[/tex] juga konvergen.
KESIMPULAN
Telah ditunjukkan bahwa [tex]\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{(n+2)(n+1)^2}\end{aligned}[/tex] konvergen, atau dengan kata lain [tex]\begin{aligned}\sum_{n=1}^{\infty}\left|\frac{1}{(n+2)(n+1)^2}\right|\end{aligned}[/tex] konvergen.
Oleh karena itu, dapat disimpulkan bahwa
[tex]\large\text{$\begin{aligned}&\sum_{n=1}^{\infty}\left[(-1)^{n+1}\frac{1}{(n+2)(n+1)^2}\right]\end{aligned}$}[/tex]
KONVERGEN MUTLAK.
