If we add the first 100 terms of the alternating series $1 - \frac { 1 } { 2 } + \frac { 1 } { 3 } - \frac { 1 } { 4 } + \frac { 1 } { 5 } - \cdots$ , how close can we determine the partial sum $S _ { 100 }$ to be to the sum $S$ of the series? A) $s _ { 100 } s \text {, with } s _ { 100 } - s s , \text { with } s _ { 100 } - s s \text {, with } s _ { 100 } - s

Question

If we add the first 100 terms of the alternating series $1 - \frac { 1 } { 2 } + \frac { 1 } { 3 } - \frac { 1 } { 4 } + \frac { 1 } { 5 } - \cdots$ , how close can we determine the partial sum $S _ { 100 }$ to be to the sum $S$ of the series? A) $s _ { 100 } > s \text {, with } s _ { 100 } - s < \frac { 1 } { 101 }$ B) $s _ { 100 } > s , \text { with } s _ { 100 } - s < \frac { 1 } { e ^ { 100 } }$ C) $s _ { 100 } > s \text {, with } s _ { 100 } - s < \frac { 1 } { 100 }$ D) $s _ { 100 } < s , \text { with } s - s _ { 100 } < \frac { 1 } { e ^ { 101 } }$ E) $s _ { 100 } > s , \text { with } s _ { 100 } - s < \frac { 1 } { e ^ { 101 } }$ F) $s _ { 100 } < s \text {, with } s - s _ { 100 } < \frac { 1 } { 101 }$ G) $s _ { 100 } < s \text {, with } s - s _ { 100 } < \frac { 1 } { 100 }$ H) $s _ { 100 } < s , \text { with } s - s _ { 100 } < \frac { 1 } { e ^ { 100 } }$

Quizplus · Accepted Answer

The series is alternating and convergent, and each term decreases in absolute value. The error in stopping at the $n$th term is less than the absolute value of the next term. Since the series is alternating, $S_{100}$ is less than $S$, and the difference $S - S_{100}$ is less than the next term, which is $\frac{1}{101}$.

If We Add the First 100 Terms of the Alternating 1−12+13−14+15−⋯1 - \frac { 1 } { 2 } + \frac { 1 } { 3 } - \frac { 1 } { 4 } + \frac { 1 } { 5 } - \cdots1−21​+31​−41​+51​−⋯

If We Add the First 100 Terms of the Alternating $1 - \frac { 1 } { 2 } + \frac { 1 } { 3 } - \frac { 1 } { 4 } + \frac { 1 } { 5 } - \cdots$