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$$ \mbox{How to show that that}\qquad \sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right) = \ln\left(3\right)\ {\large ?} $$ $$ \mbox{or}\quad 1 + \frac{1}{2} -\frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} +\frac{1}{7} + \frac{1}{8} - \frac{2}{9} \cdots = \ln\left(3\right) $$

Felix Marin
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jon jones
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4 Answers4

22

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

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

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

Ragib Zaman
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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 00:50
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    I learned something new today. Thanks, @RagibZaman. – Potato Sep 22 '12 at 17:52
8

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 $$\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}-\frac2{3n})=\ln(3N)-\ln N+O\left(\frac1N\right)=\ln3+O\left(\frac1N\right).$$

Hagen von Eitzen
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6

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.

user 1591719
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3

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

Martin Sleziak
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