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I need find a Telescopic or Geometric Series but I dont know how do that. I tried everything but nothing work. help me please

$$\sum _{ n=1 }^{ \infty }{ \frac { 2n }{ \left( n+1 \right) ! } } $$

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    $\begingroup$ Hint: We have $2n=(2n+2)-2$. $\endgroup$ – André Nicolas Feb 5 '15 at 3:56
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We have $$\sum_{n = 1}^\infty \frac{2n}{(n+1)!} = 2\sum_{n = 1}^\infty \frac{(n+1) - 1}{(n+1)!} = 2\sum_{n = 1}^\infty \left(\frac{1}{n!} - \frac{1}{(n+1)!}\right)$$

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  • $\begingroup$ Omg! I'm so stupid >.< this is a joke... Thanks you dude! $\endgroup$ – ZellAllon Feb 5 '15 at 4:00
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Hint. Since you are looking for a telescoping sum, try something like $$\frac{2n}{(n+1)!}=\frac{a}{n!}-\frac{a}{(n+1)!}\ .$$ Can you find a value of $a$ which makes this work?

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