For nЄN, find and prove a formul a for ∑1 / (i(i+1))?

Question

For nЄN, find and prove a formul a for ∑1 / (i(i+1)).

Notes: N = Natural numbers.
Also I think n^2 / n(n+1) not sure how to prove or if correct?
Sum goes from i=1 to n.

Answer 1

Please note that:

1/(k(k+1)) = 1/k - 1/(k+1).

Thus, the sum from 1 to 4 would be:

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5.

Do you see what's happening here? All the middle terms cancel each other out, leaving only the terms on the end. Therefore, [k=1, n]∑1/(k(k+1)) = 1 - 1/(n+1). This may be proven formally by induction as follows:

First, note that this holds true for n=1, since [k=1, n]∑1/(k(k+1)) = 1/2 = 1-1/2, which is what we get by plugging n into the formula. Now suppose it holds for some n -- then [k=1, n+1]∑1/(k(k+1)) = [k=1, n]∑1/(k(k+1)) + 1/((n+1)(n+2)) = 1 - 1/(n+1) + 1/((n+1)(n+2)) = 1 - 1/(n+1) + 1/(n+1) - 1/(n+2) = 1 - 1/(n+2), which is what we would get by substituting n+1 into the formula. Therefore, if it holds for some n, it must also hold for n+1, and since it holds for 1, then by induction, it holds for all natural numbers. Q.E.D.

Edit: Well, looks like I've been beaten to the punch. Of course, this also agrees with the formula you found, since n²/(n(n+1)) = n/(n+1) = 1 - 1/(n+1).

Answer 2

Your suggested answer appears to be correct. You prove it using induction.

1. Prove that it works for n=1. This is easy

2. Prove that IF it works for n = k, then it also works for n=k+1.

So Sum(1/(i(i+1))) = k²/k(k+1)

Add term for k+1:

S(k+1 terms) = k²/(k(k+1)) + 1/((k+1)(k+2)

See can you work this round to be equal to (k+1)²/((k+1)(k+2))

If so, the proposition is proved.

Answer 3

The result is easily obtained by splitting the expression into partial fractions. This gives 1/(i+1) - 1/i.
For i from 1 to n, this sums to 1/(n+1) -1 = n/(n+1).
Your answer is correct. You can cancel an n from top and bottom.

Answer 4

1/(i(i+1)) = 1/i - 1/(i+1).
It follows:
sum(from 1 to n)[1/(i(i+1))]
= 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)
= n/(n+1).
So, you are correct.

Answer 5

1/i(i+1) = 1/i - 1/(i+1)
when we add from i =1 to n they cancell out and we are left with

1 - 1/(n+1) or n/(n+1)

your solution reduces to n/(n+1) when we cancell n so you are right

step

1/1 - 1/2 + 1/2 - 1/3 ....- 1/n + 1/n - 1/(n+1) = 1- 1/(n+1)

Answer 6

You are correct, though it reduces to n/(n+1).
Can do a proof by induction. Show for n=1 or two...assume it works for n and then show it will work for n+1...gotta go...maybe I'll show later...