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+...#?

Question

Evelyn Agard · Accepted Answer

undefined To find the $ n $th partial sum of the series $ 1 - 2 + 4 - 8 + \ldots + (-2)^n + \ldots $, follow these steps:

1. Write out the general term of the series, which is $ a_n = (-2)^{n-1} $.
2. Calculate the partial sum $ S_n $ up to the $ n $th term by summing the first $ n $ terms of the series.
3. The sum of the first $ n $ terms $ S_n $ can be expressed as $ S_n = a_1 + a_2 + \ldots + a_n $.
4. Substitute the expression for the general term $ a_n $ into the partial sum $ S_n $.
5. Simplify the expression for $ S_n $ to find a formula for the $ n $th partial sum.
6. To determine whether the series converges, analyze the behavior of the partial sums as $ n $ approaches infinity.
7. If the partial sums approach a finite limit as $ n $ goes to infinity, the series converges. Otherwise, it diverges.
8. If the series converges, find its sum by evaluating the limit of the partial sums as $ n $ approaches infinity.

To find the sum when it exists, evaluate the limit of the partial sums $ S_n $ as $ n $ approaches infinity.

Emilia Aguilar · 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." # # "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." #