Finding $\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}+\cdots+\frac{1}{S_{2013}}$

Assume $S_1=1 ,S_2=1+2,S=1+2+3+,\ldots,S_n=1+2+3+\cdots+n$

How to find : $$\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}+\cdots+\frac{1}{S_{2013}}$$

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What were your difficulties in attempting the problem? – Ron Gordon Jan 18 '13 at 10:32

Hint 1:

$$\sum_{k=1}^n k = \frac{n (n+1)}{2}$$

Hint 2:

$$\frac{1}{n (n+1)} = \frac{1}{n} - \frac{1}{n+1}$$

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you can see $s_{n}=1+2+\cdots=\dfrac{n(n+1)}{2}$, so

$\dfrac{1}{s_{n}}=\dfrac{2}{n(n+1)}=2\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$

then

$\dfrac{1}{s_{1}}+\dfrac{1}{s_{2}}+\cdots+\dfrac{1}{s_{2013}}=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{2013}-\dfrac{1}{2014}\right)=\dfrac{2013}{1007}$

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$$\sum_{n=1}^{2013}\frac{1}{n(n+1)/2}=\sum_{n=1}^{2013}\frac{2}{n(n+1)}=2\sum_{n=1}^{2013}\frac{1}{n(n+1)}=2\sum_{n=1}^{2013}\left(\frac{1}{n}-\frac{1}{n+1}\right)=$$ $$2\sum_{n=1}^{2013}\frac{1}{n}-2\sum_{n=1}^{2013}\frac{1}{n+1}=2\sum_{n=1}^{2013}\frac{1}{n}-2\sum_{n=2}^{2014}\frac{1}{n}=2\sum_{n=1}^{2013}\frac{1}{n}-2\left(\sum_{n=1}^{2013}\frac{1}{n}-1+\frac{1}{2014}\right)=$$ $$=2-\frac{2}{2014}$$

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