"This is a geometric series with first term 1, and common ratio -2."
"The" \ \ n^{"th"} \ \ "partial sum, is:"
\qquad \qquad \qquad \qquad \qquad \qquad \quad \ 1 - 2 + 4 - 8 + \cdots + (-2)^n.
"Recall the formula for the sum of a finite geometric series with"
"first term 1, and common ratio" \ \ r":"
\qquad \qquad \qquad \qquad 1 + r + r^2 + r^3 + \cdots + r^n \ = \ { r^{ n + 1 } - 1} / { r - 1 }.
:. \qquad \ 1 - 2 + 4 - 8 + \cdots + (-2)^n \ = \ { (-2)^{ n + 1 } - 1} / { (-2) - 1 }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ { (-2)^{ n + 1 } - 1} / { -3 }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ { 1 - (-2)^{ n + 1 } } / { 3 }.
"Thus:"
\qquad \qquad \ 1 - 2 + 4 - 8 + \cdots + (-2)^n \ = \ { 1 - (-2)^{ n + 1 } } / { 3 }. \qquad \quad (1)
"Now we look at convergence of the infinite series:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad sum_{k=0}^{\infty} \ (-2)^k. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (2)
"We recall that if the general infinite series" \ \ sum_{k=0}^{\infty} \ a_k \ \ "converges,"
"then, among other things," \ \ lim_{k rarr \infty} a_k = 0. \ \ "So, if the desired"
"series in (2) converges, then we have, among other things:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad lim_{k rarr \infty} (-2)^k = 0.
"But the sequence" \ \ (-2)^k \ \ "clearly diverges, and by oscillation."
"Thus, the series in (2) diverges:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad sum_{k=0}^{\infty} \ (-2)^k \ \ "diverges."
"So, summing up our results (forgive the pun !), we have:"
\qquad \qquad \quad 1 - 2 + 4 - 8 + \cdots + (-2)^n \ = \ { 1 - (-2)^{ n + 1 } } / { 3 }.
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad sum_{k=0}^{\infty} \ (-2)^k \qquad \quad "diverges."
