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

Hint: First, we will identify the series. Then we will find the starting term and the common factor from the given question. Then, we will apply the formula for the sum of a finite geometric series and get the answer.

Complete step by step answer:
According to the question, we know that the given series is a geometric series. From the given series we get that the starting term will be 1 and the common factor from the series is (-2).
We get that the nth partial sum will be:
\[1 - 2 + 4 - 8 + ... + {( - 2)^n} + ...\]
We know that there is a formula for the sum of a finite geometric series that had a starting term 1 and had a common factor as ‘r’. The formula is:
\[1 + r + {r^2} + {r^3} + ... + {r^n} = \dfrac{{{r^{n + 1}} - 1}}{{r - 1}}\]
When we put the values of our question according to this formula, we get:
\[1 - 2 + 4 - 8 + ... + {( - 2)^n} = \dfrac{{{{( - 2)}^{n + 1}} - 1}}{{( - 2) - 1}}\]
Here, according to the formula, we can see that the common factor is (-2). This means r=-2.
Now, when we simplify the equation. We will open the brackets of the denominator, and we get:
\[1 - 2 + 4 - 8 + ... + {( - 2)^n} = \dfrac{{{{( - 2)}^{n + 1}} - 1}}{{ - 3}}\]
To remove the negative sign from the denominator, we will try to rearrange the numerator, and we get:
\[1 - 2 + 4 - 8 + ... + {( - 2)^n} = \dfrac{{ - 1 + {{( - 2)}^{n + 1}}}}{{ - 3}}\]
Here, we will take the negative sign of the numerator as common, and we get:
\[1 - 2 + 4 - 8 + ... + {( - 2)^n} = \dfrac{{ - (1 - {{( - 2)}^{n + 1}})}}{{ - 3}}\]
Now, the negative sign gets cancelled from both the numerator and denominator, and we get:
\[1 - 2 + 4 - 8 + ... + {( - 2)^n} = \dfrac{{(1 - {{( - 2)}^{n + 1}})}}{3}\]
So, we will assign this equation as equation (1) here.
Next, we will go to the convergence of infinite series. The convergence of the infinite series is:
\[\sum\limits_{k = 0}^\infty {{{( - 2)}^k}} \]
We will keep this as our equation (2) here.
We know that when the general infinite series which is \[\sum\limits_{k = 0}^\infty {{a_k}} \]is converging, then \[\mathop {\lim }\limits_{k \to \infty } \,{a_k} = 0\]. Now, when the equation (2) is converging, then we get:
\[\mathop {\lim }\limits_{k \to \infty } \,{( - 2)^k} = 0\]
But we know that the sequence \[{( - 2)^k}\]is diverging. So, the series in equation (2) is diverging. So, we get:
\[\sum\limits_{k = 0}^\infty {{{( - 2)}^k}} \,is\,diverging\]

Therefore, we get our result as:
\[1 - 2 + 4 - 8 + ... + {( - 2)^n} = \dfrac{{1 - {{( - 2)}^{n + 1}}}}{3}\] and \[\sum\limits_{k = 0}^\infty {{{( - 2)}^k}} \,diverges\].

Note: The formula for nth partial sum of a series \[\sum\limits_{k = 1}^\infty {{a_k}} \]. The nth partial sum \[ \Rightarrow \sum\limits_{k = 1}^\infty {{a_k}} = {a_1} + {a_2} + {a_3} + ... + {a_n}\]. We will find the nth partial sum for the telescoping series. We will simplify the series and rearrange the terms, and then we will get the answer.
\[{3^x} = 3\,and\,{3^x} = \dfrac{1}{9}\]