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.
Infinite series: $1/2 + 1/(1\cdot 2 \cdot 3) + 1/(3\cdot 4 \cdot 5) + \ldots$
3 Answers
Hint:
$$ \frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} $$
and
$$ 1 - \frac12 + \frac13 - \frac14 + \dotsb = \ln 2$$
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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6You 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. – 2012-07-18
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1You 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. – 2012-07-18
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0I downvoted because there are incorrect/nonsensical statements about subtracting divergent series, as mentioned by anon and ncmathsadist. $\infty - \infty = \ln (2)$? – 2012-07-18
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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0Use dollar signs to render $\LaTeX$. – 2018-09-17
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0The $\TeX$ 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