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.
 A: 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.
A: 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.
A: 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.
A: 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 $S=2$
A: $\small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$
$\small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$
Telescopic sum 
$1+\sum_{k=2}^{\infty} a_k =?$
