(+50): Deret Geometris, , Buktikan bahwa deret geometris di atas ekuivalen dengan 2., Sebagai tantangan, jangan gunakan rumus deret geometris untuk menyelesaikan.

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(+50): Deret Geometris


 \displaystyle\sum_{ x=0  }^{  \infty    } \left( \frac{ 1  }{ 2  ^ { x  }    }   \right)  =2
Buktikan bahwa deret geometris di atas ekuivalen dengan 2.

Sebagai tantangan, jangan gunakan rumus deret geometris untuk menyelesaikan.​

Jawab dan Penjelasan dengan langkah-langkah:

Untuk pembuktian di bawah ini, ruas persamaan saya balik terlebih dahulu.

\large\text{$\begin{aligned}&&2&=\sum\limits_{x=0}^{\infty}\left(\frac{1}{2^x}\right)\\&&{\iff}2&=\sum\limits_{x=0}^{\infty}\left2\left(\frac{1}{2\cdot2^{x}}\right)\right\\&&{\iff}2&=\sum\limits_{x=0}^{\infty}\left2\left(\frac{1}{2^{x+1}}\right)\right\\&&&.....\ \small\text{$\sum{c\cdot f(n)}=c\cdot\sum f(n)$}\\&&{\iff}2&=2\cdot\sum\limits_{x=0}^{\infty}\left(\frac{1}{2^{x+1}}\right)\\&&&.....\ \small\textsf{manipulasi indeks}\end{aligned}$}

\large\text{$\begin{aligned}&&{\iff}2&=2\cdot\sum\limits_{x=1}^{\infty}\left(\frac{1}{2^{x}}\right)\\&&{\iff}2&=nderbrace{2\bigg(nderbrace{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots}_{\bf S_n\,,\ n=\infty}\bigg)}_{\bf2S_n\,,\ n=\infty}\\&&{\iff}2&=2S_{\infty}\qquad.....(\star)\end{aligned}$}

Untuk nilai n terbatas:

\large\text{$\begin{aligned}&&S_n&=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots+\frac{1}{2^{n-1}}+\frac{1}{2^n}\\&&{\iff}2S_n&=2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots+\frac{1}{2^{n-1}}+\frac{1}{2^n}\right)\\&&&=1+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots+\frac{1}{2^{n-1}}\right)\\&&&=1+\left(S_n-\frac{1}{2^n}\right)\\&&&=1+S_n-\frac{1}{2^n}\\&&{\iff}S_n&=1-\frac{1}{2^n}\qquad.....(\star\star)\end{aligned}$}

Ketika n mencapai nilai tak hingga:

\large\text{$\begin{aligned}&&S_{\infty}&=1-\lim\limits_{n\to\,\infty}\left(\frac{1}{2^n}\right)\\&&&=1-\frac{\lim\limits_{n\to\,\infty}\:(1)}{\lim\limits_{n\to\,\infty}\:(2^n)}\\&&&.....\ \small\text{$\lim\limits_{n\to\,\infty}\:(1)=1$}\\&&&.....\ \small\text{$\lim\limits_{n\to\,\infty}\:(2^n)=\infty$}\\&&&.....\ \small\text{$\implies\lim\limits_{n\to\,\infty}\left(\frac{1}{2^n}\right)=0$}\\&&&=1-0\\&&S_{\infty}&=\bf1\end{aligned}$}

\large\text{$\begin{aligned}&\textsf{Substitusi }S_{\infty}\rightarrow(\star):\\&2=2(1)\iff2=2\\&\textsf{Ruas kiri}=\textsf{Ruas kanan}\\&\therefore\ \textsf{Persamaan $\sum\limits_{x=0}^{\infty}\left(\frac{1}{2^x}\right)=2$ terbukti.}\end{aligned}$}


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