Prove that $$\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$$
I tried to look at $$ f_n(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^n $$
And maybe taking it's derivative but it didn't work out well.
Any ideas?
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AlonAlon
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4Take $x^{3n+1}$ instead of $x^n$. – Daniel Fischer Dec 13 '14 at 16:54
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Oh I actually thought about trying it but somehow gave up on that – AlonAlon Dec 13 '14 at 16:55
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1Change the $ x^n $ into $ x^{3n} $. Then $ f (x) $ is $\frac{1}{x}$ times the integral from $0$ to $ x$ of what series? – kobe Dec 13 '14 at 16:55
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1I talk to derivatives too sometimes, it's our secret. Also think about the natural log... – GPerez Dec 13 '14 at 16:56
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@GPerez, You should try to have a conversation with an imporer integral then! – AlonAlon Dec 13 '14 at 16:57
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1I've tried but they're just so uncouth. Now that you've edited though I sound like a weird guy – GPerez Dec 13 '14 at 16:59
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1Oh and disregard the part in my first comment that actually addresses the question, I thought you could approach the problem with the $\log$ series but it seems a different solution has been shown. – GPerez Dec 13 '14 at 17:02
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2I have answered an exactly similar problem on AoPS & I give a complete answer, you may visit the link I give. – Anastasiya-Romanova 秀 Dec 13 '14 at 17:09
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@Anastasiya-Romanova秀, could you explain the second equality in your answer? (the change from summation to the integral with $e$) – AlonAlon Dec 13 '14 at 17:42
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@AlonAlon Do you mean interchanging the integral & summation sign? It can be justified by Fubini-Tonelli theorem. – Anastasiya-Romanova 秀 Dec 13 '14 at 18:02
2 Answers
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Hint
$$\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}=\lim_{N\to\infty}\sum_{n=0}^N(-1)^n\int_0^1t^{3n}dt=\lim_{N\to\infty}\int_0^1\sum_{n=0}^N(-t^3)^ndt=\lim_{N\to\infty}\int_0^1\frac{1-(-t^3)^{N+1}}{1+t^3}dt$$
Now you need to prove two things:
- $$\int_0^1\frac{dt}{1+t^3}=\frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$$
- $$\lim_{N\to\infty}\int_0^1\frac{(-t^3)^{N+1}}{1+t^3}dt=0$$
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1Which equality do you mean? Interchanging $\sum$ and $\int$? It's possible with finite sum. – Dec 13 '14 at 17:13
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1@AlonAlon Since $0<t<1$, then as $N\to\infty$ the integrand goes to $0$ – Anastasiya-Romanova 秀 Dec 13 '14 at 17:14
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Is it legal: $\lim_{N\to\infty} \int_0^1 (-t^3)^{N+1} = \int_0^1 0 = 0$? (Meaning, talking the limit before evaluating the integral) – AlonAlon Dec 13 '14 at 17:28
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2To prove the second point use that $$\left|\int_0^1\frac{(-t^3)^{N+1}}{1+t^3}dt\right|\le\int_0^1 t^{3N+3}dt=\frac1{3N+4}$$ and then squeeze. – Dec 13 '14 at 17:34
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1We can't interchange in general $\lim$ and $\int$ without theorems: Dominated convergence theorem, limit monotone theorem, convergence uniform theorem,... Or prove the result by hand as I did. – Dec 13 '14 at 17:42
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This is a very nice approach especially I like the second part where the remainder is shown as tending to $0$. This is the way such elementary problems should be be done (i.e. without any appeal to uniform convergence, dominated convergence etc). +1 – Paramanand Singh Dec 14 '14 at 04:22
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But Fubini and Tonelli's theorems justifies it for all sums/integrals less than infinity. These are less than infinity because they converge. We can directly apply that. – Amad27 Dec 26 '14 at 08:56
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For a geometric series.
$$\sum_{n=0}^{\infty} (-1)^n(x^n) = \frac{1}{1+x}$$
Substitute $x \rightarrow x^3$
$$\sum_{n=0}^{\infty} (-1)^n (x^{3n}) = \frac{1}{1+x^3}$$
Integrate the sides.
$$\sum_{n=0}^{\infty} (-1)^n\frac{x^{3n + 1}}{3n + 1} = \int \frac{dx}{1+x^3}$$
The hard part is the integration. Then let $x=1$
Amad27
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