Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 8 down vote accepted


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


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

share|cite|improve this answer

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 $\ln(2)$

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.