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Help me with this problem please.

$\arctan(x)$ can be expressed as $\displaystyle \sum_{n\ge 1}\frac{(-x)^{n-1}}{2n-1} = x-x^3/3+x^5/5-x^7/7 + \ldots$ But not sure how to proceed with the Taylor expansion. There is some extension to this problem too. Please help me how to proceed.

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    you need to drop factorials. with factorials you get $\sin x$ and not $\arctan x$. – 2017-01-19
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    oh yeah! thank you for the edit – 2017-01-19

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HINT:

The Taylor series for $f(x)=\arctan(x)$ is

$$f(x)=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$$

Now, integrate the series term by term.

To evaluate the series written in the link, combine the series for $\arctan(x)$ with the series for $\log (1+x)$ and let $x=1$.

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    :-) Looks good. Maybe want to add the expansion of $\ln(1+x)$. – 2017-01-19
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    Yeah, convincing... Well, I got the hint for series expansion... :) Thank you @Dr. MV – 2017-01-19
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We can start with $\displaystyle f(x)=\frac{1}{1+x^2}=1-x^2+x^4-\cdots$ and integrate to obtain the taylor series , $\displaystyle \arctan x = \sum_{n\ge 0}\frac{(-x)^{n+1}}{2n+1}$

Another way is to see $\displaystyle \arctan(x) = \frac{1}{2i}\left(\ln(1+ix)-\ln(1-ix)\right)$ and use the taylor expansion of $\displaystyle\ln(1+x)=\sum_{n\ge1}\frac{(-x)^{n-1}}{n}$

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    :-/ The $\arctan$ and $\ln$ complex relationship may be a bit confusing, but other than that this is a good answer. – 2017-01-19
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    That's the way how gregory derived $\theta$ in terms of $\tan \theta$ ;) – 2017-01-19
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    the relationship between ln and arctan is not confusing, but I got messed up at arctanx expressed as a summation @AdityaNarayanSharma – 2017-01-19