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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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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