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I am not sure how to find the sum of this series

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

I know it converges due to the alternating series test because the function is decreasing over its domain and the limit as $n$ approachs infinite is zero. However, I don't know what method to use to compute the finite value it converges to.

Also I know the answer is $\pi/4$ because Wolframalpha but I want to know how. Thanks.

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    $\begingroup$ Write down the Taylor series for $\arctan x$ and use $x = 1$. $\endgroup$ – user296602 Nov 6 '17 at 20:36
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This can be shown from the Taylor series

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

Letting $x=1$

$$ \sum_{n=0}^\infty (-1)^n \frac{1}{2n+1} = \arctan 1 = \frac{\pi}{4} $$

For a proof of this, infer from the geometric series $$ \frac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n} $$

Then integrate both sides

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  • $\begingroup$ Thank you! I forgot about Taylor series! $\endgroup$ – John Nov 7 '17 at 7:51
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If your interest is in computing the value (and you're interested in more alternating series than just this one), then you may be interested in series acceleration techniques. The basic method for accelerating alternating series, Euler's method, has a great 'heuristic' derivation in terms of operators:

Consider two operators: the shift operator $E$ that, when applied to the element $a_n$ of a sequence, produces the element $Ea_n = a_{n+1}$; and the difference operator $\Delta$ that gives the difference between the next and current elements, $\Delta a_n=a_{n+1}-a_n$. Then these operators satisfy the conceptual relation $E=1+\Delta$ (where strictly speaking here '$1$' is actually the identity operator).

Now, the series $\sum_n (-1)^na_n$ can be written as $\sum_n (-1)^nE^na_0$; each term in the sequence $a_n$ is obtained by applying the shift operator $E$ $n$ times to the first term. But this 'looks' like a geometric series, so we can formally write $\sum_n (-1)^na_n$ $=\sum_n(-1)^nE^na_0$ $=\frac1{1+E}a_0$. Now, substitute $E=1+\Delta$ into this to get $\sum_n(-1)^na_n = \frac1{2+\Delta}a_0$. And expand again using the geometric series : $\frac1{2+\Delta}a_0$ $= \frac12\frac1{1+\Delta/2}a_0$ $=\frac12\sum_n(-1)^n\left(\frac\Delta2\right)^na_0$ $= \sum_n\frac{(-1)^n}{2^{n+1}}\Delta^na_0$.

This is all heuristic and needs to be rigorously justified, of course, but it turns out that that justification can in fact be given; the transformed series will converge if the original series does, and converge to the same value. What's more, it will typically converge much faster; in the case of your series where $a_n=\frac1{2n+1}$ it's easy to show by induction that $\Delta^ia_n = (-1)^i\dfrac{(2\cdot 4\cdot\ldots\cdot (2i))}{(2n+1)(2n+3)\ldots(2n+2i+1)}$ and so $\Delta^i a_0 = (-1)^i\dfrac{(2\cdot4\cdot\ldots\cdot(2i))}{1\cdot3\cdot\ldots\cdot (2i+1)}$ $=(-1)^i\dfrac{2^2\cdot4^2\cdot\ldots\cdot(2i)^2}{(2i+1)!}$ $=(-1)^i\dfrac{2^{2i}(i!)^2}{(2i+1)!}$. This turns out to be roughly proportional to $i^{-1/2}$ as $i\to\infty$ and so the convergence of the 'transformed' series $\frac\pi4=\sum_n\dfrac{2^{n-1}(n!)^2}{(2n+1)!}$ is basically geometric, giving you roughly one bit of precision per term.

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