Prove that the sum of n/(n+1)! equals 1 - 1/(n+1)!

[sdrmybrat]

Prove that:
∀ n€N [(the) sum of an (infinite?) series (a1,+a2,...+,an)] (where $$a_{n}$$=$$\frac{n}{(n+1)!}$$)
$$\sum \frac{n}{(n+1)!}$$ (is equal to/gives/yields) = 1 - $$\frac{1}{(n+1)!}$$

Prove that:
∀ n $$\in$$ N $$\sum \frac{n}{(n+1)!}$$ = 1 - $$\frac{1}{(n+1)!}$$

THX in advance

 

[EnumaElish]

What is the summation index?

 

[sadhu]

how it can be
$$\sum\frac{n}{n+1!}$$=$$\sum\frac{1}{n!}\frac{1}{n+1!}$$
there is a negative sign between last two expessions in n
e-1-(e-2)=1

 

[Renmazuo]

Hey there,

A possible derivation of the sum requested uses the telescoping series property.
Note that for every j, Aj can be expended to -

Aj = j / ( j + 1 )! = 1 / j! - 1 / ( j + 1)!

Summing over 1,...,n would then yield the desired result.