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How do I calculate this: $$\frac{1}{2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}+\dots $$ I have not been sucessful to do this.

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2 Answers 2

up vote 8 down vote accepted

Hint:

$$ \frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} $$

and

$$ 1 - \frac12 + \frac13 - \frac14 + \dotsb = \ln 2$$

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You wish to find the sum $$\frac{1}{2} + \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k)(2k+1)}$$ Expanding the summand using partial fractions, we get $$\frac{1}{(2k-1)(2k)(2k+1)}=\frac{A}{2k-1}+\frac{B}{2k}+\frac{C}{2k+1}$$$$ \implies 1=A(2k)(2k+1)+B(2k+1)(2k-1)+C(2k)(2k-1)$$ Solving this gives $A=C=\frac{1}{2},B=-1$. Thus splitting up our sum, we arrive at: $$\frac{1}{2}+\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{2k-1}+\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{2k+1}-\sum_{k=1}^{\infty}\frac{1}{2k}$$ Now note that $$\sum_{k=1}^{\infty}\frac{1}{2k+1}=\sum_{k=1}^{\infty}\frac{1}{2k-1}-1$$ So our halves cancel, and grouping terms leaves us with: $$\sum_{k=1}^{\infty}\frac{1}{2k-1}-\sum_{k=1}^{\infty}\frac{1}{2k}$$ In other words, $1-\frac{1}{2}+\frac{1}{3}-\ldots$ which is known to converge to $\ln(2)$

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6  
You might want to be careful with rearrangement. Many of the series you have diverge, and so conditional convergence considerations throw the calculations into doubt. Working with partial sums would be more rigorous, and would basically parallel the argument anyway. –  anon Jul 18 '12 at 15:42
1  
You should put a finite limit on the sums, such as $N$ instead of $\infty$. You will see that the tail terms that ooze out are small and go to zero. –  ncmathsadist Jul 18 '12 at 15:45
    
I downvoted because there are incorrect/nonsensical statements about subtracting divergent series, as mentioned by anon and ncmathsadist. $\infty - \infty = \ln (2)$? –  Jonas Meyer Jul 18 '12 at 16:30

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