Sum of Series
Problem
Let n
be a positive integer. Prove that
| n |
|
n(n-1) |
|
n(n-1)...(1) |
|
| ——— |
- |
————— |
+ ... + (-1)n-1( |
————————— |
) = 1/2 |
| (n+1) |
|
(n+1)(n+2) |
|
(n+1)(n+2)...(2n) |
|
Solutions
What at seemed initially a straightforward approach -- finding a common
denominator, and showing the numerator is just half that -- turned into a real
mess. I waded through it -- see "My First Solution" below -- but
it still isn't pretty.
As it turns out, there are lots and lots of different ways to solve this one.
When the solutions were published by Dr. Alec Mihailovs (the URL appears,
below) I was amazed at their simplicity. Later, he emailed me yet another
solutions, which is listed as "Solution 3" below.
My First Solution
(The hard way)
First, a lemma.
Show that (n+1)(n+2)...(2n-1) is equal to the sum of A0 + A1 +
A2 + ... + An-2 + An-1, where
A0 = (1-n)(2-n)...(-3)(-2)(-1)
A1 = (1-n)(2-n)...(-3)(-2) * (2n)
A2 = (1-n)(2-n)...(-3) * (2n-1)(2n)
...
Ak = (1-n)(2-n)...(-k-2)(-k-1) * (2n-k+1)(2n-k+2)...(2n)
...
An-2 = (1-n) * (n+3)...(2n-1)(2n)
An-1 = (n+2)(n+3)...(2n-1)(2n)
| A0 + A1 |
= |
(1-n)(2-n)...(-3)(-2) * (2n-1) |
| A2 |
= |
(1-n)(2-n)...(-3) * (2n)(2n-1) |
| A0 + A1 + A2 |
= |
(1-n)(2-n)...(-3) * (2n-2)(2n-1) |
Continuing to add successive terms Ak in this way, we see that
| A0 + ... + Ak-1 |
= |
(1-n)(2-n)...(-k-1)(-k) * (2n-k+1)...(2n-1) |
| Ak |
= |
(1-n)(2-n)...(-k-1) * (2n)(2n-k+1)...(2n-1) |
| A0 + ... + Ak |
= |
(1-n)(2-n)...(-k-1) * (2n-k)(2n-k+1)...(2n-1) |
So the when k=n-1, the factors (1-n)(2-n)...(-k-1) on the left side of the
expression disappear, and this sum is given by
| A0 + ... + An-1 |
= |
(n+1)(n+2)...(2n-1) |
And the lemma is proved.
Now the solution itself. From the statement of the problem,
|
n |
|
n(n-1) |
|
n(n-1)...(1) |
| Let B = |
——— |
- |
—————— |
+ ... + (-1)n-1 |
———————— |
|
(n+1) |
|
(n+1)(n+2) |
|
(n+1)(n+2)...(2n) |
Factoring out the common term, n,
|
1 |
|
(n-1) |
|
(n-1)(n-2) |
|
(n-1)...(1) |
|
| B = n ( |
——— |
- |
—————— |
+ |
————————— |
+ ... + (-1)n-1 |
————————— |
) |
|
(n+1) |
|
(n+1)(n+2) |
|
(n+1)(n+2)(n+3) |
|
(n+1)(n+2)...(2n) |
|
Expressing this with a common denominator of (n+1)(n+2)...(2n),
|
(n+2)(n+3)...(2n) + (1-n)*(n+3)...(2n)
+ (1-n)(2-n)*(n+4)...(2n) + ... + (1-n)(2-n)...(-2)(-1) |
|
| B = n ( |
————————————————————————————————————————————————————— |
) |
|
(n+1)(n+2)...(2n) |
|
The sum in the numerator is exactly the sum A0+A1+...+An-1, from the lemma, above, so B can be simplified,
|
(n+1)(n+2)...(2n-1) |
|
| B = n ( |
——————————— |
) |
|
(n+1)(n+2)...(2n-1)(2n) |
|
Canceling common terms,
B = 1/2
Which solves the problem.
The Published Solutions
(Gently lifted from http://www.shepherd.edu/mathweb/solution9.html)
|
Let n be a positive integer. Prove that
|
Solution 1
|
Denote S the left hand side of the identity. Then
|
Solution 2
|
For nonnegative integers n and k, denote a(0, k) = 1 and
|
|
for positive n. Prove that
|
|
using induction by n. For n = 0 it is true. Suppose that it is true for
n--1 in place of n. Then
|
|
By the Principle of Mathematical Induction, it is true for all n. Denoting
S the left hand side of the identity in Problem 9,
|
Solution 3
Dr. Alec Mihailovs writes,
I know another interesting solution of Problem 9, by the way, using difference
operator
Delta(an) = an+1 - an
Then
(1 + Delta)an = an+1
and
(1+Delta)k an = an+k
Using the binomial formula for
(1+Delta)n-1
and applying it to the sequence
an=1/(n+1)
for which
Delta(an) = -1/((n+1)(n+2)),
Delta2(an) = 2/((n+1)(n+2)(n+3)),
Delta3(an) = -6/((n+1)(n+2)(n+3)(n+4)), etc.,
and multiplying by n at the end,
we will get the identity of Problem 9. Isn't that amazing?
Upon reading that, I was prepared to be amazed. But not being quite
so quick, I had to puzzle it out for a while...
In the formula (1+Delta)k an = an+k, we
let k=n-1, so we get
(1+Delta)n-1 an = a2n-1
Using the binomial expansion,
a2n-1 = 1 + (n-1) Delta + (1/2)(n-1)(n-2) Delta2 +
...
1/(2n) = 1 + (n-1) Delta + (1/2)(n-1)(n-2) Delta2 + ...
1/(2n) = 1 + (n-1)/((n+1)(n+2)) + (n-1)(n-2)/((n+1)(n+2)(n+3)) + ...
n/(2n) = n + n(n-1)/((n+1)(n+2)) + n(n-1)(n-2)/((n+1)(n+2)(n+3)) + ... = 1/2
Now, I'm amazed!
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