# Finding limit of a (Laurent?) series

I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one: $$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + n}$$ The task is to find the limit.

We have $$1+2+...+k=\frac{k(k+1)}2$$ So the sum you need to compute is $$\begin{split} \sum_{k=1}^n \frac 2{k(k+1)} &= \sum_{k=1}^n 2\left ( \frac 1 k - \frac 1 { k+1} \right )\\ &=2-\frac 2 {n+1}\\ &=\frac {2n} {n+1} \end{split}$$ Now you can take the limit.

Hint: Recall the sum of an arithmetic series of consecutive numbers $$1,2, \cdots, n$$: $$1+2+\cdots+n=\frac{n(n+1)}{2}$$ Take reciprocal of it and deal with partial fraction, the terms will be eliminated. $$\frac{2}{n(n+1)}=2(\frac{1}{n}-\frac{1}{n+1})$$ After that, take the limit.

Hint: Use the fact that $$1+2+...+n=\frac{n(n+1)}{2}$$

The series then becomes $$2\sum\limits_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1} \right)$$ which is a telescoping series.

As $$1+2+\ldots +n = \frac{n}{2}(n+1)$$ you are looking for the sum of the series $$S = \sum_{n=1}^{\infty} \frac{1}{\frac{k}{2}(k+1)} = 2 \sum_{k=1}^{\infty} \frac{1}{k(k+1)}$$.

Nos separating you get that $$S_n = 2\sum_{k=1}^{n} \frac{1}{k} - \frac{1}{k+1} = 2(1 - \frac{1}{n+1})$$. Now taking limit when $$n \rightarrow \infty$$ you get that $$S=2$$

$$\small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$$

$$\small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$$

Telescopic sum

$$1+\sum_{k=2}^{\infty} a_k =?$$