Finite Sum Convergence: Solving for n in the Equation n^2 + 3n - 2 = 0

[Nick Bruno]

1. Homework Statement [/b]

See attached

2. Homework Equations

See attached

3. The Attempt at a Solution

I know the answer is 6 or zero... but I can't figure out how to "show" this. When typing this equation into my calculator, I can clearly see that the number always ends in .0 or .6.

Initially I let k = (n-(n-1))

so the equation turns into

(n-(n-1))*([n-(n-1)] +1) / n = n(n-1)/n (1)

I then solve for n and I get

n^2 + 3n - 2 = 0

My calculator produces results

n^2 + 3n + 2 = 0.

Is equation (1) the correct way to solve this problem? If it is, the I just have a sign error somewhere... but I've triple checked my work and it seems there is none, so I am assuming (1) is not correct. Any suggestions?

 

[CompuChip]

Initially I let k = (n-(n-1))

I've never seen this type of question before, so I wouldn't know how to solve it right away.
But I do want to point out that n-(n-1)) = 1. So you are letting k = 1, is that what you meant to say?

By the looks of it your hunch is right though, the sum is always either an integer, or something of the form (integer) + (2/3). You might want to try finding a closed form of the result (i.e. a general formula for all n) and then consider n modulo 3...

 

[Mark44]

In the sum below, you can move the n in the denominator outside the summation.
\sum_{k = 1}^n \frac{k(k + 1)}{n}~=~\frac{1}{n}\sum_{k = 1}^n k(k + 1)

Now you can rewrite k(k + 1) as k² + k, and split the single sum into two separate summations, not forgetting the factor of 1/n for each.

Using the well-known representations for \sum{k^2} and \sum{k}, you can get a nice, neat expression for your original sum.