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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.

3 Answers 3

9

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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    I downvoted because there are incorrect/nonsensical statements about subtracting divergent series, as mentioned by anon and ncmathsadist. $\infty - \infty = \ln (2)$?2012-07-18
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This is an infinite series, but it is not geometric because there is no common ratio.

So, let $S = 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} + \dots + n \cdot \frac{1}{2^n} + \dotsb.$ Multiplying this equation by 2, we get $2S = 1 \cdot 1 + 2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{4} + \dots + n \cdot \frac{1}{2^{n - 1}} + \dotsb.$ Subtracting these equations, we find $S = 1 \cdot 1 + 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + \dots + 1 \cdot \frac{1}{2^n} + \dotsb.$ So, $S=1+\dfrac12+\dfrac14+\dfrac18+\dotsb.$ Even though it didn't begin as one, we've managed to rewrite $S$ as an infinite geometric series. Thus, we may easily find its sum: $S = \frac{1}{1 - 1/2} = \boxed{2}.$

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    The $\TeX$ delimiters `\[ ... \]` are not interpreted by MathJax. To write displayed math, use either `$ ... $` or a $\LaTeX$ environment such as `\begin{equation} ... \end{equation}`.2018-09-17