# PROOF (Sequences & Series); Can anyone help me out?

1. Jan 28, 2008

### 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 - $$\frac{1}{(n+1)!}$$

2. Jan 28, 2008

### EnumaElish

What is the summation index?

3. Jan 28, 2008

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

Last edited: Jan 28, 2008
4. Jan 30, 2008

### 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.