### Video Transcript

Find the partial sum of the series the sum from π equals one to β of five π divided by π cubed plus five π squared plus six π.

The question gives us a series, and it wants us to find an expression for the partial sum of this series. And remember, the partial sum π sub π of a series is just the sum of the first π terms of the series. So, letβs start by trying to find an expression for our πth partial sum directly. Itβs equal to the sum from π equals one to π of five π divided by π cubed plus five π squared plus six π. If we expand this series term by term, we get the following expression.

And as we start calculating this term by term, we can see a problem. Thereβs no easy way to evaluate this expression for our πth partial sum. One thing we can do when this happens is try to simplify our summand. If our summand is easier, then calculating the partial sum will also be easier. So, letβs clear some space and try simplifying our summand.

The first thing we notice is the numerator and the denominator share a factor of π. Canceling the shared factor of π gives us a new summand of five divided by π squared plus five π plus six. We might then just try calculating our partial sum directly using our new summand. However, this will still have the same problems we did before. We want to simplify this expression even further.

One thing we could try is using partial fractions. To do this, we need to notice that our numerator factors to give us π plus three times π plus two. This is because three times two is equal to six and three plus two is equal to five. So, our denominator has two unique roots. This means by using partial fractions, we can rewrite our summand five divided by π plus three times π plus two as π΄ divided by π plus two plus π΅ divided by π plus three for sum constants π΄ and π΅.

We now want to find the values of π΄ and π΅. To do this, weβll multiply both sides of this equality by π plus two times π plus three. Doing this and then simplifying, we get five is equivalent to π΄ times π plus three plus π΅ times π plus two. And remember, we use an equivalent sign because this is true for all values of π. And since this is true for all values of π, we can substitute π is equal to three to eliminate our variable π΄, this will let us find our variable π΅.

Substituting π is equal to negative three, we get five is equal to π΄ times negative three plus three plus π΅ times negative three plus two. And we can simplify this expression. Negative three plus three is equal to zero, and negative three plus two is equal to negative one. So, we get five is equal to negative π΅. Weβll then multiply both sides of this equation by negative one to get that π΅ is equal to negative five.

We now want to find the value of π΄. Weβll do this by substituting π is equal to negative two into our equation to eliminate the variable π΅. Substituting π is equal to negative two, we get five is equal to π΄ times negative two plus three plus π΅ times negative two plus two. And we can then simplify this equation. It just simplifies to give us that five is equal to π΄.

So, substituting our values of π΄ and π΅ into the expression we got for partial fractions. Weβve shown that we can rewrite our summand of five divided by π plus three times π plus two as five divided by π plus two minus five divided by π plus three. But how would this help us find the πth partial sum of our series? If we look closely at this expression, if this were the summand of a series, we can see that weβre actually getting a telescoping series. Each time we add a term into our series, we can cancel part with the term which came before.

So, letβs try using this to find an expression for the nth partial sum of our series. We have π sub π is equal to the sum from π equals one to π of five π divided by π cubed plus five π squared plus six π. Then, by canceling a shared factor of π and using partial fractions. Weβve rewritten this as the sum from π equals one to π of five divided by π plus two minus five divided by π plus three. Letβs now write this series out term by term.

The first term in our series will be when π is equal to one. This gives us five divided by one plus two minus five divided by one plus three. Weβll simplify both denominators in these expressions. We get five over three minus five over four. Next, we need to add the second term in our series when π is equal to two. We get five over two plus two minus five over two plus three. Again, weβll simplify both denominators. We get five over four minus five over five.

And now, we can see where the telescope comes in. The first part of our second term cancels with the second part of our first term. And this will be true for our entire series. For example, when we add the third term, the first part of our third term will be five divided by five. This will cancel with the second part of our second term. For our πth partial sum, we need to add terms all the way up to π is equal to π.

This term is five divided by π plus two minus five divided by π plus three. But remember, every time we add a term, the first part of that term cancels with the second part of the term which came before. So, in actual fact, five divided by π plus two cancels with the second part of the term when π is equal to π minus one. So, we only have two terms which didnβt cancel. We have five over three minus five divided by π plus three. So, this means weβve shown our πth partial sum is equal to five over three minus five divided by π plus three. And this is our final answer.

Therefore, weβve shown the πth partial sum, π sub π, of our series the sum from π equals one to β of five π divided by π cubed plus five π squared plus six π is equal to five over three minus five divided by π plus three.