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Question: Answered & Verified by Expert
The limit of $\sum_{n=1}^{1000}(-1)^{n} x^{n}$ as $x \rightarrow \infty$
MathematicsLimitsWBJEEWBJEE 2013
Options:
  • A does not exist
  • B exists and equals to 0
  • C exists and approaches to $+\infty$
  • D exists and approaches $-\infty$
Solution:
2110 Upvotes Verified Answer
The correct answer is: exists and approaches to $+\infty$
$\lim _{x \rightarrow-} \sum_{n=1}^{1000}(-1)^{n} x^{n}$
$=\lim _{x \rightarrow-}\left\{-x+x^{2}-x^{3}+x^{4}+\ldots+x^{1000}\right\}$
$=\lim _{x \rightarrow \infty}(-x) \cdot\left\{\frac{(-x)^{1000}-1}{(-x-1)}\right\}=\lim _{x \rightarrow-\infty} \frac{x^{1001}-x}{x+1}$
$=\lim _{x \rightarrow-} \frac{x^{1000}-1}{1+\left(\frac{1}{x}\right)}=+\infty$

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