Question Video: Finding the Partial Sum of a Series and Deciding Whether the Series Is Convergent or Divergent | Nagwa Question Video: Finding the Partial Sum of a Series and Deciding Whether the Series Is Convergent or Divergent | Nagwa

# Question Video: Finding the Partial Sum of a Series and Deciding Whether the Series Is Convergent or Divergent Mathematics • Higher Education

Find the partial sum for the series β_(π = 1) ^(β) 2(1/2)^(π β 1). Is the series convergent or divergent?

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### Video Transcript

Find the partial sum for the series the sum from π equals one to β of two times one-half to the power of π minus one. Is the series convergent or divergent?

We could begin by listing out the first few terms of this series. The πth partial sum is the sum of the first π terms. So, itβs π one plus π two plus π three plus π four all the way up to π sub π, where here π sub π is two times a half to the power of π minus one. And this means that π sub one is found by replacing π with one. We get two times a half to the power of one minus one, which is two times a half to the power of zero, or just two.

The second term, π sub two, is two times a half to the power of two minus one, which is two times one-half to the power of one, or just two times a half. The third term is two times a half to the power of three minus one, which is two times one-half squared. And we continue this process all the way up to the term two times a half to the power of π minus one.

Now, at first glance, it might look like this is a really complicated partial sum. But, in fact, we have a special type of series. Itβs a geometric series. Each term is obtained from the preceding one by multiplying it by the common ratio π. Now, there is a formula we can quote to find the partial sum of a geometric series. Itβs given by π times one minus π to the πth power over one minus π. So, if we can define π and π from our series, weβll be able to quite easily find the πth partial sum.

Well, we know that the first term in our series is two times a half to the power of zero. Well, a half to the power of zero is one. So, thatβs two times one. And we can, therefore, say that π is equal to two. And then, we can see that π is equal to one-half. Each time, our term is being multiplied by one-half. And so, substituting these values of π and π into the formula for the πth partial sum, and we get two times one minus a half to the πth power over one minus a half.

Now, in fact, the denominator of this fraction, one minus a half, becomes one-half. And, of course, dividing by one-half is actually the same as multiplying by two. So, we can multiply the numerator of our expression by two. And weβre left with the πth partial sum. Itβs four times one minus a half to the πth power.

The second part of this question asks us to decide whether this series is convergent or divergent. Well, in fact, we could find the limit as π approaches β of π sub π. Alternatively, we can quote a general result. We say that a geometric series with a common ratio of π is convergent if the absolute value of π is less than one. And itβs divergent if the absolute value of π is greater than or equal to one.

Well, here, π is equal to one-half. And the absolute value of one-half is one-half, which is less than one. And so, in this case, we can say that this series is convergent. In fact, weβre also able to generalise this and say that if the series is convergent, itβs sum is given by π over one minus π. In this case, thatβs two over one minus one-half, which is equal to four.

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