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$$\lim _{n\to \infty }\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots+\left(-1\right)^{n-1}\cdot \frac{1}{2n-1}\right)=\text{ ?}$$ I don't know how we get to find the result of this operation...

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3 Answers 3

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Or $$\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{2n-1} = \sum_{n=1}^\infty \int_0^1 (-1)^{n-1}x^{2n-2}\,dx= \int_0^1 \sum_{n=1}^\infty (-x^2)^{n-1}dx= \int_{0}^1 \frac{1}{1+x^2}\,dx= \cdots\dfrac{\pi}{4}$$

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  • $\begingroup$ How you obtain the second equality? $\endgroup$
    – Lucas
    Mar 29, 2015 at 0:56
  • $\begingroup$ and $\sum _{n=1}^{\infty }\left(-x^2\right)^{n-1}=\:\frac{1}{1+x^2}$ ? how you do that? $\endgroup$
    – Lucas
    Mar 29, 2015 at 1:01
  • $\begingroup$ heyy dude, if I put every value from sum don't will be 1/1-u... I verify $\endgroup$
    – Lucas
    Mar 29, 2015 at 1:08
  • $\begingroup$ I've right was a geometric series, but your formula is for general case? cause I obtain $\frac{1-u^n}{1-u}$ $\endgroup$
    – Lucas
    Mar 29, 2015 at 1:14
  • $\begingroup$ You got to take the limit: $\dfrac{1-u^n}{1-u} \to \dfrac{1}{1-u}$ $\endgroup$
    – DeepSea
    Mar 29, 2015 at 1:23
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Recall that the Taylor series for $\tan^{-1}x$ is: $$x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$

It follows that the given limit is equal to exactly $\tan^{-1}1=\frac{\pi}{4}$

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  • $\begingroup$ In my book this is solve with integral... I am a student, I never work with Taylor series $\endgroup$
    – Lucas
    Mar 29, 2015 at 0:49
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Taylor series expansion of $\displaystyle\tan^{-1}(x)$ is:

$\displaystyle\tan^{-1}(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\ldots+\left(-1\right)^{n}\frac{x^{2n+1}}{2n+1}= \sum_{n=0}^\infty (-1)^{n} \cdot \frac{x^{2n+1}}{2n+1}$

Limit you are trying to calculate is simply $\displaystyle\lim_{n\to \infty }\tan^{-1}(1)$.

Hence, $\displaystyle\lim _{n\to \infty }\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...+\left(-1\right)^{n-1}\cdot \frac{1}{2n-1}\right) = \frac{\pi}{4}$

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  • $\begingroup$ But without Taylor series, because I don't work with this, I am a student!!! $\endgroup$
    – Lucas
    Mar 29, 2015 at 1:34
  • $\begingroup$ All answers are based on Taylor series, give me an example without this method with a recurrence relation, something else $\endgroup$
    – Lucas
    Mar 29, 2015 at 1:35
  • $\begingroup$ @Lucas do you at least know the integral? Then it can be explained without Taylor $\endgroup$
    – Vim
    Mar 29, 2015 at 4:04
  • $\begingroup$ Vim if you know something else to make clever with me very good, look here and help math.stackexchange.com/posts/1210854/edit , otherwise shut up, thank you :) $\endgroup$
    – Lucas
    Mar 29, 2015 at 7:26
  • $\begingroup$ @Lucas. Please don't make assumptions. Well I don't mean to make clever with you. I was just saying that it could be explained without Taylor , since you didn't know that , but required the knowledge of this integral. BTW, if you want to reply to my comment, you should at least @ me in your comment. Otherwise I will never ever get informed. $\endgroup$
    – Vim
    Mar 29, 2015 at 10:07

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