Question

I need to prove this calculus question. It's about the sum of an alternating series. (see comment)

Answer 1

welcome to openstudy @tabby2014

Answer 2

\[\frac{ \pi }{ 6 } = \frac{ 1 }{\sqrt{3}} \sum_{n=0}^{\infty} (-1)^n \frac{ 1 }{ 3^n(2n+1) }\]

Answer 3

let me call some help @tabby2014 @Preetha

Answer 4

@satellite73

Answer 5

It looks like you want to find a function \(f(x)\) such that the integral of that function is the sum you have. The first thing I would do, is write out a few of the first values of the sum. \[a_0=1\]\[a_1=-\frac{1}{9}\]\[a_2=\frac{1}{45}\]\[a_3=-\frac{1}{189}\]

Answer 6

Now, you want to find a function \(f(x)\) such that \[f(0)=1\]\[f'(0)=-\frac{1}{9}\]and so on to\[f^{(n)}(0)=(-1)^n\frac{1}{3^n(2n+1)}\]

Answer 7

How would you even begin to do that?

Answer 8

@KingGeorge isn't
\[\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}\] the expansion of arctangent at one? not sure that helps, but it looks related

Answer 9

That's a good observation. I'll have to think about this a little more then.

Answer 10

that at least will give you \(\frac{\pi}{4}\) so probably on the right track

Answer 11

In any case, the method I was using is used for something slightly different. I don't think it would work here.

Answer 12

Is this applying the alternating series test?

Answer 13

I think I might be stuck on this. I also have to leave. I'll be back later to take a look at any progress/with new ideas.

Answer 14

Alright. Just had a brilliant idea, though I wouldn't have seen it without @satellite73's observation, and bit of research on different power series.

First, recall that\[\tan^{-1}(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}.\]Then, if we substitute \(\sqrt x\) for \(x\), we get \[\tan^{-1}(\sqrt x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{n+1/2}.\]Finally, divide by \(\sqrt x\) to get\[\frac{\tan^{-1}(\sqrt x)}{\sqrt x}=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{n}.\]If we put \(x=1/3\), we get exactly the series you put above.

Answer 15

Then of course, you divide by \(\sqrt3\) to get the exact same thing, but that's easy.