Question

find a formula for the $n^{th}$ partial sum $s_n$ of the following series. if the series is finite, determine the sum. if the series is infinite, determine if it converges or diverges, and if it converges, determine the sum. please simplify your solution. $sum_{i = 1}^{infty}(1^i - 1^{i - 1})$

Explanation:

Step1: Expand the partial sum

$$\begin{align*} S_{n}&=\sum_{i = 1}^{n}(1^{i}-1^{i - 1})\\ &=(1^{1}-1^{0})+(1^{2}-1^{1})+(1^{3}-1^{2})+\cdots+(1^{n}-1^{n - 1}) \end{align*}$$

Step2: Simplify by cancellation

When we expand the sum, we can see that most of the terms cancel out.
\(S_{n}=1^{n}-1^{0}\)

Since \(1^{n}=1\) for all \(n\geq1\) and \(1^{0} = 1\), we have \(S_{n}=1 - 1=0\) for all \(n\geq1\)

Step3: Analyze the infinite - series

For the infinite series \(\sum_{i = 1}^{\infty}(1^{i}-1^{i - 1})\), we know that \(\lim_{n
ightarrow\infty}S_{n}=\lim_{n
ightarrow\infty}(1^{n}-1^{0})\)

Since \(\lim_{n
ightarrow\infty}1^{n}=1\), \(\lim_{n
ightarrow\infty}S_{n}=1 - 1=0\)

Answer:

\(S_{n}=0\), converges, sum: \(0\)