Video: Finding the Partial Sum for a Given Series and Determining Whether It Is Convergent or Divergent

Find the partial sum of the series βˆ‘_(𝑛 = 1) ^(∞) 2/((𝑛 + 3)(𝑛 + 4)). Is the series convergent or divergent?

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Video Transcript

Find the partial sum of the series the sum from 𝑛 equals one to ∞ of two divided by 𝑛 plus three times 𝑛 plus four. Is the series convergent or divergent?

The question wants us to find an expression for the partial sum of the series. It then wants us to determine if this series is convergent or divergent. We recall that a partial sum 𝑆 π‘˜ is the sum of the first π‘˜ terms of our series. So this gives us that our π‘˜th partial sum, 𝑆 sub π‘˜, is equal to the sum from 𝑛 equals one to π‘˜ of two divided by 𝑛 plus three times 𝑛 plus four. And we notice that we can simplify our summand by using partial fractions.

Since we have two distinct factors in our denominator, we can write two divided by 𝑛 plus three times 𝑛 plus four as 𝐴 divided by 𝑛 plus three plus 𝐡 divided by 𝑛 plus four for some constants 𝐴 and 𝐡. Next, we multiply both sides of this equation by the denominator 𝑛 plus three times 𝑛 plus four. This gives us two is equal to 𝐴 times 𝑛 plus four plus 𝐡 times 𝑛 plus three. In fact, this will be true for all values of 𝑛. We represent this by using an equivalent sign.

Since this is true for all values of 𝑛, we can eliminate our variables by using substitution. Substituting 𝑛 is equal to negative four gives us that two is equal to zero minus 𝐡. Therefore, multiplying both sides of our equation by negative one gives us that negative two is equal to 𝐡. Similarly, we can eliminate the variable 𝐡 from our equation by substituting 𝑛 is equal to negative three. Substituting 𝑛 is equal to negative three and then simplifying gives us that 𝐴 is equal to two.

So we’ve shown by partial fractions that two divided by 𝑛 plus three times 𝑛 plus four is equal to two divided by 𝑛 plus three minus two divided by 𝑛 plus four. So we can use this to simplify the summand in our partial sum. So by using partial fractions, we’ve shown that our π‘˜th partial sum is equal to the sum from 𝑛 equals one to π‘˜ of two divided by 𝑛 plus three minus two divided by 𝑛 plus four.

We can now evaluate this by writing it out term by term. To get our first term, we substitute 𝑛 is equal to one. Then we add the second term where 𝑛 is equal to two. And we do this all the way up to the π‘˜th term, where 𝑛 is equal to π‘˜. Simplifying the denominators in our first term gives us two over four minus two over five. Simplifying the denominators in our second term gives us two over five minus two over six. And we see that the negative two over five in our first term and the two over five in our second term cancel. We would see a similar story in our third term. The negative two over six cancels with the positive two over six.

And now we see a pattern. Every time we add a new term, it cancels with the last part of the previous term. We call a series with this property a telescopic series. So when we added our final term of two divided by π‘˜ plus three minus two divided by π‘˜ plus four, the first part canceled with the last part of the previous term. By simplifying our first term to a half, we’ve shown that the π‘˜th partial sum is equal to a half minus two divided by π‘˜ plus four.

We can simplify this into one fraction by using cross-multiplication. This gives us one times π‘˜ plus four minus two times two all divided by two times π‘˜ plus four. And we see that our numerator simplifies to just give us π‘˜. So we’ve shown that the π‘˜th partial sum is equal to π‘˜ divided by two times π‘˜ plus four.

In fact, we don’t have to use the π‘˜th partial sum. We could use the 𝑛th partial sum, giving us that the 𝑛th partial sum, 𝑆 𝑛, is equal to 𝑛 divided by two times 𝑛 plus four.

The next part of the question is to determine whether our series is convergent or divergent. And we recall that we call a series convergent if the limit as 𝑛 approaches ∞ of the 𝑛th partial sum is equal to some finite value 𝐿. Otherwise, we call the series divergent.

So to check if our series is convergent or divergent, we need to check the limit as 𝑛 approaches ∞ of the 𝑛th partial sum. We’ve shown that this is equal to the limit as 𝑛 approaches ∞ of 𝑛 divided by two times 𝑛 plus four. Since the factor of one-half is constant with respect to 𝑛, we can take it outside of our limit.

There are then several different ways of evaluating this limit. We could use algebraic division to divide 𝑛 by 𝑛 plus four. Or we could divide by the highest power of 𝑛 in our numerator and our denominator. The highest power of 𝑛 in our numerator or our denominator is just 𝑛 itself. So we’ll divide both of these through by 𝑛. Dividing our numerator of 𝑛 by 𝑛 just gives us one. We can then divide each term in our denominator by 𝑛. This gives us one plus four divided by 𝑛.

Now we see that our numerator of one remains constant as 𝑛 is approaching ∞. In our denominator, the one remains constant as 𝑛 is approaching ∞. However, four divided by 𝑛 approaches zero. That means this limit is approaching one, which means we can evaluate the limit as 𝑛 approaches ∞ of our 𝑛th partial sum to be one-half.

Therefore, since the limit of our 𝑛th partial sum approached a finite value of one-half, we can conclude that our series is convergent. Therefore, by finding the partial sum of the series the sum from 𝑛 equals one to ∞ of two divided by 𝑛 plus three times 𝑛 plus four. We’ve shown not only is this series convergent, but we’ve shown that it converges to the value of one-half.

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