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# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given 1-2+4-8+...+(-2)^n+...?

Feb 19, 2018

#### Answer:

$\setminus q \quad \setminus q \quad \setminus \quad 1 - 2 + 4 - 8 + \setminus \cdots + {\left(- 2\right)}^{n} \setminus = \setminus \frac{1 - {\left(- 2\right)}^{n + 1}}{3} .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus {\sum}_{k = 0}^{\setminus \infty} \setminus {\left(- 2\right)}^{k} \setminus q \quad \setminus \quad \text{diverges.}$

#### Explanation:

$\text{This is a geometric series with first term 1, and common ratio -2.}$

$\text{The" \ \ n^{"th"} \ \ "partial sum, is:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus 1 - 2 + 4 - 8 + \setminus \cdots + {\left(- 2\right)}^{n} .$

$\text{Recall the formula for the sum of a finite geometric series with}$
$\text{first term 1, and common ratio" \ \ r":}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 1 + r + {r}^{2} + {r}^{3} + \setminus \cdots + {r}^{n} \setminus = \setminus \frac{{r}^{n + 1} - 1}{r - 1} .$

$\therefore \setminus q \quad \setminus 1 - 2 + 4 - 8 + \setminus \cdots + {\left(- 2\right)}^{n} \setminus = \setminus \frac{{\left(- 2\right)}^{n + 1} - 1}{\left(- 2\right) - 1}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus \frac{{\left(- 2\right)}^{n + 1} - 1}{- 3}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus \frac{1 - {\left(- 2\right)}^{n + 1}}{3} .$

$\text{Thus:}$

$\setminus q \quad \setminus q \quad \setminus 1 - 2 + 4 - 8 + \setminus \cdots + {\left(- 2\right)}^{n} \setminus = \setminus \frac{1 - {\left(- 2\right)}^{n + 1}}{3} . \setminus q \quad \setminus \quad \left(1\right)$

$\text{Now we look at convergence of the infinite series:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {\sum}_{k = 0}^{\setminus \infty} \setminus {\left(- 2\right)}^{k} . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(2\right)$

$\text{We recall that if the general infinite series" \ \ sum_{k=0}^{\infty} \ a_k \ \ "converges,}$
$\text{then, among other things," \ \ lim_{k rarr \infty} a_k = 0. \ \ "So, if the desired}$
$\text{series in (2) converges, then we have, among other things:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {\lim}_{k \rightarrow \setminus \infty} {\left(- 2\right)}^{k} = 0.$

$\text{But the sequence" \ \ (-2)^k \ \ "clearly diverges, and by oscillation.}$
$\text{Thus, the series in (2) diverges:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {\sum}_{k = 0}^{\setminus \infty} \setminus {\left(- 2\right)}^{k} \setminus \setminus \text{diverges.}$

$\text{So, summing up our results (forgive the pun !), we have:}$

$\setminus q \quad \setminus q \quad \setminus \quad 1 - 2 + 4 - 8 + \setminus \cdots + {\left(- 2\right)}^{n} \setminus = \setminus \frac{1 - {\left(- 2\right)}^{n + 1}}{3} .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {\sum}_{k = 0}^{\setminus \infty} \setminus {\left(- 2\right)}^{k} \setminus q \quad \setminus \quad \text{diverges.}$