Help Finding a Pattern in Fractions: 1/2, 1/6, 1/12, 1/20, 1/30, 1/42, 1/56

[1+1=1]

Is there anyone who can help me? I need to find a pattern in these numbers:
1/2, 1/6, 1/12, 1/20, 1/30, 1/42, 1/56. Now, I know just by looking at the denominators, if I could only work with those, I could use the formula:
2n+a_n_1. But I have that fraction, so it's all screwy. Anyone see where I'm screwing up? I know I can't.

 

[TD]

How about u_n = \frac{1}<br /> {{n\left( {n + 1} \right)}}

 

[1+1=1]

well, this is a summation problem. it is the summation of 1/n(n+1) equals "what." the "what" is what I'm supposed to find. the 1/n(n+1) is already given for the one side of the equation, i need to find what the summation is also equal to.

 

[gnpatterson]

have you try just doing a few terms by hand, it might give you a clue!

 

[TD]

I thought you were looking for a pattern to find the formula.
Are you looking for a formula for a partial sum or for the sum of the infinite series?

 

[1+1=1]

have you try just doing a few terms by hand, it might give you a clue!

Do you honestly think I have not tried that. Give me some credit.

 

[TD]

Have you thought of the expansion?

\frac{1}<br /> {{n\left( {n + 1} \right)}} = \frac{1}<br /> {n} - \frac{1}<br /> {{n + 1}}

 

[PrinceOfDarkness]

1/n(n+1)=1/n - 1/(n+1)

If you write the sum: Sn= (1/2-1/6) + (1/6-1/12) + ... + (1/n - 1/(n+1) )
All the terms cancel except 1/2 and 1/n+1
i.e. Sn= 1/2 - 1/(n+1)= (n-1)/2(n+1)

Taking the limit n->infinity, we get: 1/2

I hope it's the correct answer!

 

[TD]

The first element (n = 1) already is 1/2 so it has to be more. Other than that, your work looks good so it should be 1/2 + your 1/2 = 1

The partial sum is

s_n = \frac{n}<br /> {{n + 1}}

So for the infinite series

\mathop {\lim }\limits_{n \to \infty } s_n = \mathop {\lim }\limits_{n \to \infty } \frac{n}<br /> {{n + 1}} = 1

 

[PrinceOfDarkness]

You are right. I made a mistake. The first term of Sn should've been '1', since 1/n gives 1 for n=1.

So the series converges to 1, and all terms from 1/2 to 1/n cancel in the partial sum to give, Sn=n/(n+1), just as you point out.