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How to show that that $\sum_{n=1}^{\infty} \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}= \ln(3)$? or $1+\frac{1}{2} -\frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} +\frac{1}{7} + \frac{1}{8} - \frac{2}{9} \dotsb = \ln(3)$

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4 Answers

up vote 10 down vote accepted

Use the fact that $ \displaystyle \frac{1}{m+1} = \int^1_0 x^m dx$ to get $$ \sum_{n=1}^{\infty} \int^1_0 \left( x^{3n-2} + x^{3n-3} -2 x^{3n-1}\right) dx$$

$$ = \int^1_0 \sum_{n=1}^{\infty} \left( x^{3n-2} + x^{3n-3} -2 x^{3n-1}\right) dx$$

$$= \int^1_0 \frac{ x+1-2x^2}{1-x^3} dx= \int^1_0 \frac{2x+1}{x^2+x+1} dx = \log (x^2+x+1) \biggr|^1_0=\log 3.$$

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How in the world did you ever think of this. –  chharvey Sep 16 '12 at 23:30
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@TestSubject528491 It is a trick worth keeping in mind. This problem has many variants (sums of 1/(kn+m) where each sum would diverge individually but the cancel out enough to converge) and this method can be used on them all, which is the main reason I've remembered it, but it is also used to find expressions for generating functions/power series. I can't think of great examples off the top of my head sorry but here's another thread where it comes up. –  Ragib Zaman Sep 17 '12 at 0:50
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I learned something new today. Thanks, @RagibZaman. –  Potato Sep 22 '12 at 17:52
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Let $s_N=\sum_{n=1}^N \frac1n$ be the $n$th partial sum of the harmonic series. Then $$\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}-\frac2{3n})=\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}+\frac1{3n})-\sum_{n=1}^N\frac1n=s_{3N}-s_N.$$ It is well-known that $s_N=\ln N +\gamma+O\left(\frac1N\right)$, hence $$\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}-\frac2{3n})=\ln(3N)-\ln N+O\left(\frac1N\right)=\ln3+O\left(\frac1N\right).$$

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Let's try one line proof using Riemann sums. So, our limit may be written as

$$\lim_{n\to\infty}\sum_{k=1}^{n} \left(\frac1{3k-2}+\frac1{3k-1}+\frac1{3k}-\frac1k\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{2n}\frac1{1+\frac{k}{n}}=\int_0^2\frac{1}{1+x}=\ln3.$$

Q.E.D.

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Let $H_n=\sum_{k=1}^n \frac1k$ denote the $n$-th harmonic number.

It is well known that $H_n \sim \ln n + \gamma$, where $\gamma$ is Euler-Mascheroni constant. (To be more precise, $\lim\limits_{n\to\infty} (H_n-\ln n) = \gamma$, which is what we actually need below.)

You are interested in $$\sum_{k=1}^n \left(\frac1{3k-2}+\frac1{3k-1}+\frac1{3k}-\frac3{3k}\right) = H_{3n}-H_n \sim \ln{3n}-\ln n = \ln 3.$$

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