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How do I calculate the first few partial sums for the Leibniz Formula? I have to do a total of ten can someone post how to calculate the first few, I'm a bit lost.

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

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    $\begingroup$ Like, $1$, then $1-\frac13 = \frac23$, then $1-\frac13+\frac15 = \frac{13}{15}$? $\endgroup$ – Daniel Fischer Mar 1 '14 at 18:49
  • $\begingroup$ Of course, to calculate $\pi$, you have to multiply the partial sums by $4$... $\endgroup$ – Thomas Andrews Mar 1 '14 at 19:15
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The $N$th partial sum of a series $\sum_{n=0}^\infty a_n$ is defined to be $$S_N=\sum_{n=0}^Na_n=a_0+a_1+\cdots+a_N.$$ (see Wikipedia). Thus the first few partial sums of the series $$\sum_{n=0}^\infty(-1)^n\frac{1}{2n+1}$$ are $$\begin{align*} S_0 & =\left[(-1)^0\frac{1}{2\cdot 0+1}\right]=1\\\\\\ S_1 & =\left[(-1)^0\frac{1}{2\cdot 0+1}\right]+\left[(-1)^1\frac{1}{2\cdot 1+1}\right]=1-\frac{1}{3}=\frac{2}{3}\\\\\\ S_2 & =\left[(-1)^0\frac{1}{2\cdot 0+1}\right]+\left[(-1)^1\frac{1}{2\cdot 1+1}\right]+\left[(-1)^2\frac{1}{2\cdot 2+1}\right]=1-\frac{1}{3}+\frac{1}{5}=\frac{13}{15}\\ \end{align*}$$ I leave it to you calculate them up to $S_{10}$ (you will probably want to use Wolfram Alpha or a calculator.)

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To calculate the sum of the series:

We know that the power series of $\frac1{1+x}$ is $$\frac1{1+x}=\sum_{n=0}^\infty (-1)^nx^n $$ so we integrate term by term for $|x|<1$ $$\arctan x=\int_0^x\frac{dt}{1+t^2}=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} $$

the series is convergent for $x=1$ then we find

$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}=\arctan 1=\frac\pi4$$

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    $\begingroup$ That was not the question asked. $\endgroup$ – Thomas Andrews Mar 1 '14 at 19:16
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Answering 4 year later! I wanted to see if there was a closed nice form for the partial sums and this post is about as good as anywhere to place my findings.

$$S_k=\sum_{n=0}^k{\frac{(-1)^n}{2n+1}}$$

$S_0=1,\\ S_1=\frac{2}{3},\\ S_2=\frac{13}{15},\\ S_3= \frac{76}{105},\\S_4=\frac{789}{945},\\ S_5=\frac{7734}{10395},\\ S_6=\frac{110937}{135135} ,\\ S_7=\frac{1528920}{2027025},\\ S_8=\frac{28018665}{34459425},\\ S_{9}=\frac{497895210}{654729075}$

And in general,

$S_n =\frac{a_n}{b_n}$

Where $a_n$ is A024199$(n)$ and $b_n=(2n+1)!!$ is A001147$(n)$ and double factorial is defined in the classic way. Based on how A024199 is defined, it doesn't look like there is a nice way to express the fraction $S_n$ without invoking $\Sigma$ notation.

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  • $\begingroup$ Downvote? I will delete if I get a second downvote. I can't quite see how this post hurts this site... Care to explain? $\endgroup$ – Mason Aug 11 '18 at 4:11

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