Solving Telescoping Series with n Terms

[7C0A0A5]

I have no idea how to do this :'( really the only part I don't understand is the ending part...like for

the infinite sum of (1/n(n+1))....I know you start of by partial fractions...then you just plug in a few numbers for n.

so I end up with:

(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))

then in another problem...the infinite sum of ( 2/{(n-1)(n+1)} ) ends up like...

(1 - 1/3) + (1/2 - 1/4) +1/3 - 1/5) + ... + (1/(n-3) - 1/(n-1)) + (1/(n-2) - 1/n)


I want to know how you get the end results...the ones with "n" in them...I don't understand how to get those numbers...or why they are what they are...

I understand the first part without the "n"...but I don't know how to end it with the "n"...I hope this makes since.

 

[NateTG]

OK, let's take a look at:
\sum_{n=1}^{4}\frac{1}{n}-\frac{1}{n+1}
So if we write it out longhand it turns into:
(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})
but we can regroup:
\frac{1}{1}+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+(-\frac{1}{4}+\frac{1}{4})-\frac{1}{5}
This is just moving the parens around, but the paired numbers nicely all zero out, leaving you with
\frac{1}{1}-\frac{1}{5}

It should be easy to see that the pattern can be extended to any number of terms so that you end up with
\sum_{n=1}^{k}\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{k+1}

There are more formal approaches to this, but I doubt that they will be particularly helpful to you.

 

[M98Ranger]

Hi there,

I understand that telescoping series can be confusing, but don't worry, with some practice, you'll get the hang of it! Let's break down the steps for solving telescoping series with n terms.

1. Write out the series: The first step is to write out the series with n terms. For example, the series (1/n(n+1)) would be written as:

(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))

2. Simplify the fractions: Next, you want to simplify the fractions within the series. In the example above, you can see that the fractions (1/2, 1/3, 1/4, etc.) are all decreasing by 1 in the denominator. This is the key to solving telescoping series. You want to manipulate the fractions so that they all have the same denominator.

3. Use partial fractions: In order to manipulate the fractions, we need to use partial fractions. This involves breaking down a fraction into smaller, simpler fractions. For the series (1/n(n+1)), we can use partial fractions to break it down into:

1/n - 1/(n+1)

4. Plug in values for n: Now that we have simplified the fractions, we can start plugging in values for n. In the first term, n = 1, so we get:

1/1 - 1/2 = 1/2

In the second term, n = 2, so we get:

1/2 - 1/3 = 1/6

As you can see, the denominator of the second term is now 6, which is the same as the denominator of the first term (2 x 3 = 6). This is the key to solving telescoping series - getting the fractions to cancel out.

5. Continue plugging in values for n: You can continue plugging in values for n until you reach the last term, which will be:

1/n - 1/(n+1)

For the last term, n = n, so we get:

1/n - 1/(n+1) = 1/n - 1/(n+1)

As you can see, the fractions cancel out, leaving us with the final result of