Exercise:
Compute, $\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{5}}}{{{2}^{i}}}}$
Solution: we
have $\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{5}}}{{{2}^{i}}}}=\sum\limits_{i=0}^{\infty }{\frac{{{\left( i+1
\right)}^{5}}}{{{2}^{i+1}}}=\frac{1}{2}\sum\limits_{i=0}^{\infty
}{\frac{{{\left( i+1 \right)}^{5}}}{{{2}^{i}}}}}$
now by using
the binomial expand for ${{\left( i+1
\right)}^{5}}={{i}^{5}}+5{{i}^{4}}+10{{i}^{3}}+10{{i}^{2}}+5i+1$
So $\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{5}}}{{{2}^{i}}}=\frac{1}{2}\left( \sum\limits_{i=0}^{\infty }{\frac{{{i}^{5}}}{{{2}^{i}}}+5\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{4}}}{{{2}^{i}}}+10\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{3}}}{{{2}^{i}}}+10\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{2}}}{{{2}^{i}}}+5\sum\limits_{i=0}^{\infty
}{\frac{i}{{{2}^{i}}}+\sum\limits_{i=0}^{\infty }{\frac{1}{{{2}^{i}}}}}}}}}
\right)}$
But \[ \sum\limits_{i=0}^{\infty
}{\frac{1}{{{2}^{i}}}=1+\sum\limits_{i=1}^{\infty }{{{\left( \frac{1}{2}
\right)}^{i}}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=2}}\]
& \[\sum\limits_{i=0}^{\infty
}{\frac{i}{{{2}^{i}}}=0+\sum\limits_{i=1}^{\infty }{\frac{i}{{{2}^{i}}}}}=2 \]
since $\sum\limits_{i=1}^{\infty
}{\frac{i}{{{2}^{i}}}=\sum\limits_{i=0}^{\infty
}{\frac{i+1}{{{2}^{i+1}}}=}\frac{1}{2}\left( \sum\limits_{i=0}^{\infty
}{\frac{i}{{{2}^{i}}}+\sum\limits_{i=0}^{\infty }{\frac{1}{{{2}^{i}}}}}
\right)=\frac{1}{2}\left( \sum\limits_{i=1}^{\infty }{\frac{i}{{{2}^{i}}}+2}
\right)}$ $\Rightarrow \left( 1-\frac{1}{2} \right)\sum\limits_{i=1}^{\infty
}{\frac{i}{{{2}^{i}}}=1}$ $\Rightarrow \sum\limits_{i=1}^{\infty
}{\frac{i}{{{2}^{i}}}}=2$
Thus $\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{2}}}{{{2}^{i}}}}=\frac{1}{2}\left( \sum\limits_{i=1}^{\infty
}{\frac{{{i}^{2}}}{{{2}^{i}}}+4+2} \right)=\frac{1}{2}\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{2}}}{{{2}^{i}}}+3}$ $\Rightarrow \left( 1-\frac{1}{2}
\right)\sum\limits_{i=1}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}=3\Leftrightarrow
\sum\limits_{i=1}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}=6}}$
Same work we
will work for $\sum\limits_{i=0}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}}\,\,\And
\,\,\sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}}$
$\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{3}}}{{{2}^{i}}}=0+\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{3}}}{{{2}^{i}}}=\sum\limits_{i=0}^{\infty }{\frac{{{\left( i+1
\right)}^{3}}}{{{2}^{i+1}}}=\frac{1}{2}\left( \sum\limits_{i=0}^{\infty
}{\frac{{{i}^{3}}}{{{2}^{i}}}+3\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{2}}}{{{2}^{i}}}+3\sum\limits_{i=0}^{\infty
}{\frac{i}{{{2}^{i}}}+\sum\limits_{i=0}^{\infty }{\frac{1}{{{2}^{i}}}}}}}
\right)}}}$
$=\frac{1}{2}\left(
\sum\limits_{i=1}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}+3\left( 6
\right)+3\left( 2 \right)+2} \right)$ $\Rightarrow \sum\limits_{i=1}^{\infty
}{\frac{{{i}^{3}}}{{{2}^{i}}}}=\frac{1}{2}\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{3}}}{{{2}^{i}}}+13}$ $\Rightarrow \sum\limits_{i=0}^{\infty
}{\frac{{{i}^{3}}}{{{2}^{i}}}=26}$
Similarly we
expand $\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{4}}}{{{2}^{i}}}=0+\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{4}}}{{{2}^{i}}}=\sum\limits_{i=0}^{\infty }{\frac{{{\left( i+1
\right)}^{4}}}{{{2}^{i+1}}}}}}=\frac{1}{2}\sum\limits_{i=0}^{\infty
}{\frac{{{\left( i+1 \right)}^{4}}}{{{2}^{i}}}}$
$\Rightarrow
\sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}}=\frac{1}{2}\left(
\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{4}}}{{{2}^{i}}}+4\sum\limits_{i=0}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}+6\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{2}}}{{{2}^{i}}}+4\sum\limits_{i=0}^{\infty
}{\frac{i}{{{2}^{i}}}+\sum\limits_{i=0}^{\infty }{\frac{1}{{{2}^{i}}}}}}}}
\right)$
$=\frac{1}{2}\sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}+\frac{1}{2}\left(
4\left( 26 \right)+6\left( 6 \right)+4\left( 2 \right)+2 \right)}$ $\Rightarrow
\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{4}}}{{{2}^{i}}}}=\frac{1}{2}\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{4}}}{{{2}^{i}}}+75}$ $\Leftrightarrow \sum\limits_{i=0}^{\infty
}{\frac{{{i}^{4}}}{{{2}^{i}}}=150}$
Thus $\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{5}}}{{{2}^{i}}}=\frac{1}{2}\sum\limits_{i=0}^{\infty
}{\frac{{{i}^{5}}}{{{2}^{i}}}+\frac{1}{2}\left( 5\left( 150 \right)+10\left( 26
\right)+10\left( 6 \right)+5\left( 2 \right)+2 \right)}}$ thus $\sum\limits_{i=1}^{\infty
}{\frac{{{i}^{5}}}{{{2}^{i}}}=1082}$