<s>[INST] Calculate $\sum\limits_{n\geq 1}\frac{1}{1+2+\ldots+n} $ Calculate  $\displaystyle \sum_{n\geq 1}\dfrac{1}{1+2+\ldots+n} $ 
My attempts :
\begin{aligned}
 \sum_{n\geq 1}\dfrac{1}{1+2+\ldots+n} &= \sum_{n\geq 1}\dfrac{2}{n(n+1)}=2\sum_{n\geq 1}\left( \dfrac{1}{n}-\dfrac{1}{n+1}\right)=2\left( \dfrac{1}{1}-\dfrac{1}{+\infty}\right)=2(1-0)=2?
\end{aligned}
 [/INST] \begin{aligned}
\sum_{n\geq 1}\dfrac{2}{n(n+1)}=2\sum_{n\geq 1}\dfrac{1}{n}-2\sum_{n\geq 1}\dfrac{1}{n+1}
\end{aligned}
False. The equality does not hold because the two sums on the right are divergent, while the sum on the left is convergent.

Better go at it this way:
$$\sum_{n=1}^\infty\frac{1}{n(n+1)} = \sum_{n=1}^\infty\frac{1}{n}-\frac{1}{n+1} = \left(\frac 11 - \frac12\right) + \left(\frac12-\frac13\right) + \left(\frac13-\frac14\right) + \cdots$$
Now, remember that $$\sum_{n=1}^\infty a_n = \lim_{N\to\infty} \sum_{n=1}^N a_n$$
In your case, $$\sum_{n=1}^N \frac 1n-\frac{1}{n+1}$$ can be calculated relatively easily.
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