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The function $f(x) = \frac{1}{(1+x)}$ has the Maclaurin series:

$$f(x) = \sum_{n=0}^\infty (-x)^n$$ Find the Interval of Convergence

I've gotten to $$\lim_{n\to \infty} \left\lvert \frac{(-x)^{n+1}}{(-x)^n} \right\rvert$$

Unsure where to go from here. I've viewed a few other threads and youtube videos but none seemed to help my case.

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  • $\begingroup$ What do you require about that limit to guarantee convergence of the series? $\endgroup$
    – Umberto P.
    Commented Sep 29, 2016 at 1:20
  • $\begingroup$ Question states it converges( doesn't ask for proof ) its just a question from First year maths $\endgroup$
    – Dreeww
    Commented Sep 29, 2016 at 1:21
  • $\begingroup$ I'm not sure what that means. $\endgroup$
    – Umberto P.
    Commented Sep 29, 2016 at 1:23
  • $\begingroup$ Edited to exactly what my professor asked. $\endgroup$
    – Dreeww
    Commented Sep 29, 2016 at 1:26

2 Answers 2

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Hint: For what values of $r$ can you conclude that $$ \sum_{k=0}^{\infty}r^k=\frac{1}{1-r} $$ edit: this is circular after OP edited in what professor asked.

By the ratio test, which you started doing, you need $$ \lim_{n\rightarrow \infty}|\frac{(-x)^{n+1}}{(-x)^n}|<1\Rightarrow \lim_{n\rightarrow \infty}|-x|<1\Rightarrow |x|<1 $$

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For your computation...

First you need to understand $$ \frac{(-x)^{n+1}}{(-x)^n} = (-x) $$ When you see this left-hand-side, you should automatically know to convert it to the right-hand-side.

Since you did not do that, my guess is: you need to review your algebra before you take calculus.

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