Proof by induction (summation)

In summary, the homework statement is that if you sum up the limits of a sequence of numbers, the limit of the sum will be equal to the sum of the limits of the individual numbers in the sequence.
  • #1
marksyncm
100
5

Homework Statement



Prove by induction that ##\sum\limits_{k=1}^{2n} \frac{1}{k(k+1)} = \frac{2n}{2n+1}##

2. The attempt at a solution

First I showed that it is true for ##n=1## and ##n=2##. Then, assuming it is true for all ##n##, I attempt to show that it is true for ##n+1##:

$$\sum\limits_{k=1}^{2(n+1)} \frac{1}{k(k+1)} = \frac{2(n+1)}{2(n+1)+1} = ... $$

My problem is what the right side (##...##) of the equation should be. I can't just say that it is equal to ##\frac{2n}{2n+1} + \frac{1}{(n+1)(n+2)}##, because the summation notation requires me to double my ##n## - I understand this to mean that each time we add 1 to ##n##, we are adding two additional terms to the summation total - one is ##2(n+1)-1## and the other is ##2(n+1)##. Based on this reasoning, I concluded that the above equation should be:

$$\sum\limits_{k=1}^{2(n+1)} \frac{1}{k(k+1)} = \frac{2(n+1)}{2(n+1)+1} = \frac{2n}{2n+1} + \frac{1}{(2n+1)(2n+2)} + \frac{1}{(2n+2)(2n+3)}$$

However, I was unable to reduce the right side to resemble the left side, and online algebra tools were of no help. Am I going about this wrong?
 
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  • #2
I believe this is the source of your problems:
Look at the expression $$\frac{2n}{2n+1}$$. Now replace ##n## by ##(n+1)##. That gives you $$\frac{2(n+1)}{2(n+1)+1} = \frac{2n+2}{2n+3}$$

You had ##(2n+2)/(2n+4)##.

If you're still uncertain what's going on, consider this. The general expression ##2n/(2n+1)## is an even divided by an odd. Your expression is an even divided by an even.
 
  • #3
Sorry, this was actually a problem in my LaTeX - I have now corrected this. In my calculations, I had used ##\frac{2(n+1)}{2(n+1)+1}## and was unable to get the right side of the equation to resemble this, even using Mathway.
 
  • #4
That doesn't line up with the rest of your calculations, as that denominator is not ##2(n+2) = 2n+4##. Look at the last line of your original question. What is the denominator on the left hand side?

You can indeed show that if you combine your three fractions over the common denominator ##(2n + 1)(2n + 2)(2n + 3)##, the numerator reduces to ##(2n + 1)(2n + 2)^2## and so the fraction reduces to ##(2n + 2)/(2n+3)##, the desired result.
 
  • #5
RPinPA said:
That doesn't line up with the rest of your calculations, as that denominator is not ##2(n+2) = 2n+4##. Look at the last line of your original question. What is the denominator on the left hand side?

Apologies, I'm not sure I see which part you are referring to. I'm not sure I see where it still says ##2n+4## in a denominator in my original question?
 
  • #6
marksyncm said:
Apologies, I'm not sure I see which part you are referring to. I'm not sure I see where it still says ##2n+4## in a denominator in my original question?

Ah, OK, it's been edited. I was still seeing ##2(n+2)## when I wrote that. So once more, your left hand side can be rewritten as (2n + 2)/(2n + 3) and it is indeed possible and not too messy to reduce the right-hand side to that form.
 
  • #7
Ok, thank you - now that I know that my approach is correct, I can keep cracking on the reduction. Appreciated!
 

Related to Proof by induction (summation)

1. What is proof by induction (summation)?

Proof by induction (summation) is a mathematical proof technique used to prove that a statement holds for all natural numbers. It involves two steps: the base case, where the statement is proven to hold for the first natural number, and the induction step, where it is shown that if the statement holds for one natural number, it also holds for the next natural number. This process is repeated until the statement is proven to hold for all natural numbers.

2. How is proof by induction (summation) different from other proof techniques?

Proof by induction (summation) is different from other proof techniques in that it specifically deals with proving statements for all natural numbers. Other proof techniques may focus on specific cases or involve different methods, such as direct proof or proof by contradiction.

3. What are the key components of a proof by induction (summation)?

The key components of a proof by induction (summation) are the base case, the induction hypothesis, and the induction step. The base case establishes that the statement holds for the first natural number. The induction hypothesis assumes that the statement holds for a specific natural number, and the induction step shows that if the statement holds for that number, it also holds for the next natural number. These components are repeated until the statement is proven to hold for all natural numbers.

4. What types of statements can be proven using proof by induction (summation)?

Proof by induction (summation) can be used to prove statements about natural numbers, such as equations, inequalities, and divisibility. It can also be used to prove properties of sequences and series.

5. Are there any limitations to using proof by induction (summation)?

While proof by induction (summation) is a powerful and commonly used proof technique, it does have some limitations. It can only be used for proving statements about natural numbers, and it may not be the most efficient or practical method for some proofs. Additionally, it requires a clear understanding of the base case and induction step, which can be challenging in some cases.

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