PMI (bridge to abstract mathematicsds) prove ∑ from i=1 to n of 2/(i(i+1)) = n/n+1

[rochelle]

PMI (bridge to abstract mathematicsds) prove ∑ from i=1 to n of 2/(i(i+1)) = n/n+1

I have a question on this problem:
Use Principle of Mathematical Induction (PMI) to prove that for all natural numbers n,
∑ from i=1 to n of 2/(i(i+1)) = n/n+1.

so the first step of this problem is to show that the statement is true for n=1 right? but when i plug in 1 i get 1=1/2 which is false so i don't know how to proceed...

thanks in advance!

 

[ksdhart2]

Hrm... there appears to be an error somewhere in the problem. Did you perhaps accidentally copy it down wrong, or was it wrong in the text itself (wouldn't be unheard of, sadly)? I see two possible places the problem could have gone wrong. One, the summation formula is messed up and it's meant to say:

\(\displaystyle \displaystyle \sum_{i=1}^{n} \: \frac{1}{i(i+1)} = \frac{n}{n+1}\)

Or two, the result of the summation formula could be messed up and it's meant to say:

\(\displaystyle \displaystyle \sum_{i=1}^{n} \: \frac{2}{i(i+1)} = \frac{2n}{n+1}\)

Either way, what you've done seems correct. It's just the problem that's flawed.

 

[HallsofIvy]

I have a question on this problem:
Use Principle of Mathematical Induction (PMI) to prove that for all natural numbers n,
∑ from i=1 to n of 2/(i(i+1)) = n/n+1.

so the first step of this problem is to show that the statement is true for n=1 right? but when i plug in 1 i get 1=1/2 which is false so i don't know how to proceed...

thanks in advance!

Rather than induction, which requires that we already have the formula, we can use "partial fractions" to simplify the sum: \(\displaystyle \frac{2}{i(i+1)}= \frac{A}{i}+ \frac{B}{i+1}\) for some numbers A and B. Multiply both sides by i(i+1) to get \(\displaystyle 2= A(i+1)+ Bi\). This has to be true for all x. Taking i= 0 2= A. Taking i= -1, 2i= B(-1) so B= -2. That is, \(\displaystyle \frac{2}{i(i+1)}= \frac{2}{i}- \frac{2}{i+ 1}\).

Now look at the first few terms in the sum: with i= 1, that is 2- 1. With i= 2, that is 1- 2/3. Adding, the "1"s cancels leaving 2- 2/3. With i= 3, that is 2/3- 1/2. Adding that, the "2/3"s cancel leaving 2- 1/2. This is what is called a "telescoping sequence". Each term has a term that cancels a previous one. Adding up to i= n, the last term is \(\displaystyle \frac{2}{n}- \frac{2}{n+ 1}\). The first term will cancel with the previous i= n-1 leaving a total of \(\displaystyle 2- \frac{2}{n+1}= 2\left(1- \frac{1}{n+1}\right)= 2\left(\frac{n+ 1}{n+ 1}- \frac{1}{n+ 1}\right)= \frac{2n}{n+ 1}\), as ksdhart suggests.

 

[rochelle]

Yes it seems my professor made a mistake so now i am working on
\(\displaystyle \displaystyle \sum_{i=1}^{n} \: \frac{1}{i(i+1)} = \frac{n}{n+1}\)
so now i understand the first step but I'm a little stuck on the second part trying to prove by hypothesis of induction. how do i do this?

Hrm... there appears to be an error somewhere in the problem. Did you perhaps accidentally copy it down wrong, or was it wrong in the text itself (wouldn't be unheard of, sadly)? I see two possible places the problem could have gone wrong. One, the summation formula is messed up and it's meant to say:

\(\displaystyle \displaystyle \sum_{i=1}^{n} \: \frac{1}{i(i+1)} = \frac{n}{n+1}\)

Or two, the result of the summation formula could be messed up and it's meant to say:

\(\displaystyle \displaystyle \sum_{i=1}^{n} \: \frac{2}{i(i+1)} = \frac{2n}{n+1}\)

Either way, what you've done seems correct. It's just the problem that's flawed.

 

[JeffM]

Yes it seems my professor made a mistake so now i am working on
\(\displaystyle \displaystyle \sum_{i=1}^{n} \: \frac{1}{i(i+1)} = \frac{n}{n+1}\)
so now i understand the first step but I'm a little stuck on the second part trying to prove by hypothesis of induction. how do i do this?

Here is the structure

\(\displaystyle \displaystyle n = 1 \implies \sum_{i=1}^n \dfrac{1}{i(i + 1)} = \dfrac{1}{1(1 + 1)} = \dfrac{1}{2} = \dfrac{n}{n + 1}.\)

\(\displaystyle \therefore \exists\ k \in \mathbb N^+ \text { such that } \displaystyle \sum_{i=1}^k \dfrac{1}{i(i + 1)} = \dfrac{k}{k + 1}.\)

Do you see how we get that?

Now we need to prove it is true for k + 1 assuming nothing except what is specified above about k (which does NOT say that k = 1.)

The most common way to proceed is to set up the expression for k + 1
and break it down into an expression in k and an expression in k + 1.

\(\displaystyle \displaystyle \sum_{i=1}^{k+1} \dfrac{1}{i(i + 1)} = \left ( \sum_{i=1}^k \dfrac{1}{i(i + 1)} \right ) + \dfrac{1}{(k+1)\{(k + 1) + 1\}} =\\ \dfrac{k}{(k + 1)} + \dfrac{1}{(k+1)\{(k + 1) + 1\}} = what?\)

Can you take it to the end now?