9
$\begingroup$

What is the value of the sum of the series

$\frac{1}{1}+\frac{1}{2+3}+\frac{1}{4+5+6}+\dotso\;?$

And this:

$\frac{1}{1}+\frac{1}{2\cdot3}+\frac{1}{4\cdot5\cdot6}+\dotso\;?$

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    FWIW: the denominators of your second series are in the [OEIS](http://oeis.org/A057003); the series itself does not seem to have a neat closed form.2011-11-03

2 Answers 2

6

Note $a+(a+1)+\cdots+(b-1)+b= (b-a+1)(b+a)/2$. Set $b=n(n+1)/2,a=n(n-1)/2+1$: $(\circ)=\sum_{n=1}^\infty \left((n(n+1)/2-[n(n-1)/2+1]+1)\cdot\frac{n(n+1)/2+[n(n-1)/2+1]}{2}\right)^{-1}$ $=\sum_{n=1}^\infty\frac{2}{n(n^2+1)}=\sum_{n=1}^\infty\left(\frac{2}{n}-\frac{1}{n+i}-\frac{1}{n-i}\right)=\sum_{n=1}^\infty\int_0^12x^{n-1}-x^{n+i-1}-x^{n-i-1}dx$ $=\int_0^1\left(\sum_{n=1}^\infty x^{n-1}\right)(2-x^i-x^{-i})dx=\int_0^1\frac{1-x^i}{1-x}dx+\int_0^1\frac{1-x^{-i}}{1-x}dx$ $=\psi(1+i)+\psi(1-i)+2\gamma.$ (Where $\psi$ is the digamma function.) That's the first series. Second one doesn't look so easy.

3

There is a simple formula for $\sum_{i=k}^n i = \frac{n(n+1)}{2}-\frac{(k-1)k}{2}$.

Now, the m-th denominator has $m$ terms, which means that there are $1+2+..+(m-1)=\frac{(m-1)m}{2}$ terms before the first of the denominator.

Hence, your n'th term is

$\frac{1}{\sum_{i=\frac{(n-1)n}{2}+1}^{\frac{(n-1)n}{2}+n} i}$

Use the formula at the begining of the proof, and you'll probably end up with a standard telescopic sum....

Edit See Andre's comment below, this is not telescopic, so read the above comments as "how to reduce this series to a simpler "closed" form series".. The exact formula is calculated below by Andre.

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    @Eric Naslund: Yes. Exotic is relative, to you it is roughly as familiar as $x^2$.2011-11-02