# value of $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\cdots$?

Is there anything known about the value of the series $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\frac{1}{1+2+3+4+5+6}+\cdots$ ?

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Yes, in fact, everything is known about it. Step 1: get a formula for the denominators (I assume the 6th one is supposed to have a 6 in it, not stop at 5 like the 5th one). Step 2: apply partial fractions to get a telescoping series. Enjoy.

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thanks! fixed the typo after the 5yh term. –  Arjang Feb 9 '12 at 11:15
It equals 2.... –  Matthew Doucette Feb 9 '12 at 14:41
@Gerry I think the last word was meant to be "profit". –  Pedro Tamaroff Apr 18 '12 at 19:19
The denominators $a_n$ are $$a_n = \sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$ so $$\sum_{n=1}^{\infty} \frac1{a_n} = \sum_{n=1}^{\infty} \frac{2}{n(n+1)} = 2\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 2$$ If you don't have the resources to check this on a computer (out to infinity), you will have your computer or calculator carry out the calculation for $n<N$. You should then get a number near the partial sum $$s_N=\sum_{n=1}^{N-1} \frac1{a_n} = 2 \left( 1 - \frac{1}{N} \right),$$ where whatever difference you find between the computer's answer and this expected answer will be the result of truncation or rounding errors due to the machine representation of each floating point number (which usually involve binary mantissas as in IEEE formats) encountered along the way.