What is the value of the sum of the series


And this:


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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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.


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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    $\begingroup$ 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
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    $\begingroup$ @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

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