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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+...#?

1 Answer
Feb 19, 2018

Answer:

# \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." #

Explanation:

# "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." #