How is this a telescoping series?

In summary, the conversation discusses a series that needs to be evaluated when approaching infinity. It is mentioned that this is a telescopic series and the limit is 1/2. However, the person does not see how this is possible and breaks down the series into two parts. They then attempt to plug in a value for k to understand the algebra, but are unsure how to account for the negative sign. The expert advises against using intuition and encourages using algebra to solve the problem.
  • #1
arhzz
260
52
Homework Statement
Finding the limit of the series
Relevant Equations
Telescoping series formula
Hello !

Consider this series;

$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ Now if we input k to let's say 3 we get this

$$ \frac{1}{2} + \frac{1}{3} +\frac{1}{3} + \frac{1}{5} +\frac{1}{7} + \frac{1}{7} $$. I think its fair to assume that only 1/2 should remain and the others should cancel out,that is kind of the thing by the telescoping series.But I have + across the board and the fractions won't cancel out but rather add up.I am not sure how the - is supposed to come into play here,since as I've been taught yestarteday we have to sum up the elements.The only logical solution would be that within the parantheses one element has to be negative,but I don't see how we get to that.
 
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  • #2
arhzz said:
$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$
Are you sure about that?
 
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  • #3
Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
 
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  • #4
Vanadium 50 said:
Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
Okay so you are definately right,these are not the same so the algebra doesn't add up.I supposed it would have to be 1st fraction - 2nd fraction but I assumed it would be + because I don't know intuition? I saw * in the denominator and simply assumed it would be +
 
  • #5
Don't use intuition. Use algebra.
 
  • #6
Vanadium 50 said:
Don't use intuition. Use algebra.
Yea I guess I rushed it. It all makes sense now,thanks for pointing out the mistake.
 

1. What is a telescoping series?

A telescoping series is a type of infinite series where most of the terms cancel each other out, leaving only a finite number of terms to be added. This results in a simpler form of the series that can be easily evaluated.

2. How can you tell if a series is telescoping?

A series is considered telescoping if it can be written in the form of (a1-a2) + (a2-a3) + (a3-a4) + ... + (an-1-an) + an, where the terms in parentheses cancel each other out. This can also be written as a1 + (a2-a2) + (a3-a3) + ... + (an-an) + an, where the terms in parentheses add up to zero.

3. What is the significance of a telescoping series?

Telescoping series are significant because they allow for the easy evaluation of infinite series that would otherwise be difficult to solve. They also demonstrate the concept of cancellation in mathematics.

4. How do you find the sum of a telescoping series?

To find the sum of a telescoping series, you can use the formula S = a1 + (a2-a2) + (a3-a3) + ... + (an-an) + an, where S is the sum of the series and an is the last term in the series. This formula works because all the terms in parentheses cancel out, leaving only a1 and an to be added.

5. Can any series be written as a telescoping series?

No, not all series can be written as telescoping series. A series must have terms that cancel each other out in order to be considered telescoping. If the terms do not have this property, then the series cannot be written in the form of (a1-a2) + (a2-a3) + (a3-a4) + ... + (an-1-an) + an.

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