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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.

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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.

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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.

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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.

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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 $S=2$

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

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

Telescopic sum

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

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