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I have the following alternating sum: $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}$$

Or we can write it as $\sum_{n=0}^\infty \frac{(-1)^{n}}{2n+1}$ if one is more comfortable with working from this form.

I already tested this series for convergence via the alternating sums test, and I plugged it into wolframalpha just to check what it should evaluate to, and it turned up $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1} = \frac{\pi}{4}$$

My guess is the goal should be to convert the series from alternating to geometric or some form that can be easily evaluated, but I don't know how to rewrite this. It isn't just finding a way around the alternating term, I should want a term like $x^n$ but we have $\frac{1}{2n-1}$. The difference in the denominator is also something we would need to work around.

EDIT:

FROM HERE it looks like I could replace $\frac{1}{2n-1}$ and get $$\frac{1}{2n-1} = \int_0^1 x^{2n}$$ and that would give me: $$-\sum_{n=1}^\infty \int_0^1 (-x^2)^n = -\int_0^1 \sum_{n=1}^\infty (-1)^n x^{2n} = -\int_0^1 \frac{1}{1+x^2}$$

Which is technically wrong because this evaluates to $\frac{-\pi}{4}$ when it should be positive. I pulled out a negative sign from $(-1)^{n+1} = -(-1)^n$, so I'm not sure why this results in the wrong sign.

Anyways, this looks like it would handle both aspects of my problem. This approach feels rather advanced though, is this the more orthodox method of finding a way to rewrite these series when the denominator is like this?

Researcher R
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    $\frac {\pi} 4$ cannot appear from a much simpler argument, can it? – geetha290krm Dec 18 '23 at 08:24
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    See this https://artofproblemsolving.com/community/c7h1925653p13212478 – NadiKeUssPar Dec 18 '23 at 08:31
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    $$\frac{1}{2n-1} \ne \int_0^1 x^{2n}\ dx$$ – jjagmath Dec 18 '23 at 12:17
  • @jjagmath oops. So the sum starting at n = 0 would be more appropriate – Researcher R Dec 18 '23 at 22:28
  • @geetha290krm The purpose of my question was to look for a universal approach for dealing with alternating series. The approach I found is very convenient and much easier to apply than THIS where they broke up the sum into 2 sums. I guess I'm being paranoid because I haven't come across many alternating series to evaluate thus far aside from this one, so I worried that this simple approach I found might not work all the time. – Researcher R Dec 18 '23 at 22:40

1 Answers1

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First question.

The sum corresponds to the Taylor series of the arctangent function, namely
$$ \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} = \arctan x, $$ hence the wanted result when evaluating at $x = 1$.

Second question.

Note that $$ \int_0^1 x^{2n} \,\mathrm{d}x = \frac{1}{2n\color{red}{+}1}, $$ hence $$ \begin{align} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1} &= \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \\ &= \sum_{n=0}^\infty (-1)^n \int_0^1 x^{2n} \,\mathrm{d}x = \frac{1}{2n+1} \\ &= \int_0^1 \frac{\mathrm{d}x}{1+x^2} \\ &= [\arctan x]_0^1 \\ &= \frac{\pi}{4} \end{align} $$

Abezhiko
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  • Identifying the series as a Taylor series expansion of arctan was very illuminating and helps out greatly. Also, thank you for putting the mistake in red to improve visibility of it! – Researcher R Dec 18 '23 at 22:45