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:

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

$\endgroup$
  • 1
    $\begingroup$ Do you always add one more term in the denominator? In which case the next term would be $\frac{1}{7+8+9+10}$? $\endgroup$ – Olivier Bégassat Nov 2 '11 at 19:26
  • 21
    $\begingroup$ There are two close votes. I vote against closing. The question was merely formulated ungrammatically; it's a real question and the grammar has been fixed in the meantime. $\endgroup$ – joriki Nov 2 '11 at 19:28
  • 2
    $\begingroup$ FWIW: the denominators of your second series are in the OEIS; the series itself does not seem to have a neat closed form. $\endgroup$ – J. M. is a poor mathematician Nov 3 '11 at 0:33
6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ For the first question, we are looking at $\sum\frac{2}{n(n^2+1)}$. There is an exotic closed form, but no easy telescoping. $\endgroup$ – André Nicolas Nov 2 '11 at 19:34
  • $\begingroup$ It was clear that we should end up with a cubic in the denominator, I was expecting omne which is a product of three linear (mainly because I was expecting this to be an exam/assignment type question), but too lazy to calculate it. $\endgroup$ – N. S. Nov 2 '11 at 19:41
  • 1
    $\begingroup$ @AndréNicolas: You can write it nicely in terms of the logarithmic derivative of the gamma function. is this the exotic form? $\endgroup$ – Eric Naslund Nov 2 '11 at 19:57
  • $\begingroup$ @Eric Naslund: Yes. Exotic is relative, to you it is roughly as familiar as $x^2$. $\endgroup$ – André Nicolas Nov 2 '11 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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