Video: Evaluating the Partial Sums Sequence for a Given Series

Evaluate the sequence of partial sums 𝑠_(π‘˜) for the series βˆ‘_(𝑛 = 1)^(∞) 2/(𝑛(𝑛 + 1)).

03:52

Video Transcript

Evaluate the sequence of partial sums 𝑠 π‘˜ for the series the sum from 𝑛 equals one to ∞ of two divided by 𝑛 times 𝑛 plus one.

The question wants us to evaluate the sequence of partial sums for this series. And we recall that the partial sum 𝑠 π‘˜ of a series is a finite sum of the first π‘˜-terms. In our case, 𝑠 π‘˜ is given by the sum from 𝑛 equals one to π‘˜ of two divided by 𝑛 times 𝑛 plus one. We notice that the series given to us in the question has a summand of two divided by 𝑛 times 𝑛 plus one. We can manipulate this into two separate fractions by using partial fractions.

Since our denominator has two unique factors, by using partial fractions, we can write two divided by 𝑛 times 𝑛 plus one as 𝐴 divided by 𝑛 plus 𝐡 divided by 𝑛 plus one for some constants 𝐴 and 𝐡. Now we multiply both sides of the equation by the denominator 𝑛 times 𝑛 plus one. This gives us that two is equal to 𝐴 times 𝑛 plus one plus 𝐡 times 𝑛.

In fact, these will be true for any value of 𝑛. We represent this by using an equivalent sign. Even though originally our values of 𝑛 had to be positive integers, by using partial fractions, we’ll find two variants which are the same for all values of 𝑛, not just the positive integers. So we’ll use substitution to eliminate one of our variables.

We’ll start with 𝑛 is equal to zero. This gives us that two is equal to 𝐴 times zero plus one, which is just equal to 𝐴. Similarly, we can substitute 𝑛 is equal to negative one to eliminate the variable 𝐴 from our equation. Substituting 𝑛 is equal to negative one gives us that two is equal to 𝐡 multiplied by negative one. And we can multiply both sides of this equation by negative one to see that negative two is equal to 𝐡.

So by using partial fractions, we’ve shown that two divided by 𝑛 times 𝑛 plus one is equal to two divided by 𝑛 minus two divided by 𝑛 plus one. We can then use this to rewrite the partial sum of our series. By using partial fractions, we’ve shown that the π‘˜th partial sum of our series is given by the sum from 𝑛 equals one to π‘˜ of two divided by 𝑛 minus two divided by 𝑛 plus one.

We’re now ready to evaluate the π‘˜th partial sum of our series. We’ll do this by writing out our partial sum term by term. For our first term, we use 𝑛 is equal to one. This gives us two divided by one minus two divided by one plus one. For our second term, we use 𝑛 is equal to two. This gives us two over two minus two divided by two plus one. And we can see that we have negative two divided by two and positive two divided by two. So these terms cancel. And if we add in our third term, we can see we have a negative two over three and a positive two over three. And these can cancel again.

In fact, we can see from our summand that whenever we have a negative two divided by 𝑛 plus one, if we add another term, we will have two divided by 𝑛 plus one minus two divided by 𝑛 plus two. We call this type of series a telescoping series. And we can see the only parts of our partial sum that won’t cancel is the first part of our first term and the last part of our last term.

So since we’re calculating the π‘˜th partial sum, our last term will have 𝑛 equal to π‘˜. And our two divided by π‘˜ will cancel with a negative two divided by π‘˜ in the previous term. So we have that 𝑠 π‘˜ is equal to two minus two divided by π‘˜ plus one. We can write this as one fraction by multiplying two by π‘˜ plus one divided by π‘˜ plus one. Distributing two over the parentheses gives us two π‘˜ plus two. We then subtract two to get two π‘˜. This gives us two π‘˜ divided by π‘˜ plus one.

Therefore, we’ve shown that the π‘˜th partial sum for the series the sum from 𝑛 equals one to ∞ of two divided by 𝑛 times 𝑛 plus one is given by two π‘˜ divided by π‘˜ plus one.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.