Prove sequence is bounded above

[csc2iffy]

Homework Statement

Let a_n = 1 + 1/(1*2) + 1/(2*3) + ... + 1/(n*[n+1]). Prove {a_n} is bounded above.

Homework Equations

1/(2*3) = 1/2 - 1/3

The Attempt at a Solution

I accidentally left my notebook at school and I have no idea how to do this without my class notes. The book doesn't have any examples to help me out either. Please help...

 

[genericusrnme]

I'd prove the series converges and is monotonically increasing

 

[Dick]

It's a telescoping series, isn't it?

 

[Joffan]

Use your "relevant equation" on the first five or six terms and see where Dick's term comes from...

 

[csc2iffy]

ok so this is what i have so far, I'm not sure if it "proves" it completely though:
a1 = 1/1 - 1/2 = 1/(1*2)
a2 = 1/2 - 1/3 = 1/(2*3)
a3 = 1/3 - 1/4 = 1/(3*4)
...
a(n-1) = 1/(n-1) - 1/n = 1/((n-1)*n)
an = 1/n - 1/(n+1) = 1/(n*(n+1))

Adding everything together
(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/(n-1) - 1/n) + (1/n - 1/(n+1))
1 - 1/(n+1) = 1/(1*2) + 1/(2*3) + ... + 1/(n*(n+1)) = an - 1
Add 1 to both sides
an = 2 - 1/(n+1) = (2n+1)/(n+1)
lim(an) = 2
Therefore the series is bounded above by 2

does this complete my proof?

 

[Dick]

ok so this is what i have so far, I'm not sure if it "proves" it completely though:
a1 = 1/1 - 1/2 = 1/(1*2)
a2 = 1/2 - 1/3 = 1/(2*3)
a3 = 1/3 - 1/4 = 1/(3*4)
...
a(n-1) = 1/(n-1) - 1/n = 1/((n-1)*n)
an = 1/n - 1/(n+1) = 1/(n*(n+1))

Adding everything together
(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/(n-1) - 1/n) + (1/n - 1/(n+1))
1 - 1/(n+1) = 1/(1*2) + 1/(2*3) + ... + 1/(n*(n+1)) = an - 1
Add 1 to both sides
an = 2 - 1/(n+1) = (2n+1)/(n+1)
lim(an) = 2
Therefore the series is bounded above by 2

does this complete my proof?

I think that would do it, yes!

 

[Joffan]

Would it be overkill to prove a_{n}=2-\frac{1}{n+1} by induction?

Additionally my "quick and dirty" bounded-above argument would have been to observe that the terms are less than the reciprocals of squares (1/(2*3) < 1/(2*2) etc). Getting the least upper bound is a bonus, of course.