Express the series negative a half plus one-quarter minus one-eighth plus one sixteenth minus and then the series continues in sigma notation.
This notation at the end of the question here is the Greek letter Σ. And it represents a sum. We can express a series using sigma notation. In general, we write the sum from 𝑟 equals 𝑎 to 𝑏 of 𝑢𝑟. Here, 𝑢𝑟 means the general term in the series, the term with term number 𝑟. The sigma notation tells us that we are summing the terms in this series.
The values of 𝑎 and 𝑏 give the limit over which are ranges. The value of 𝑎 gives the number of the first term that we’re summing. And the value of 𝑏 gives the number of the last term in the sum. For example, if 𝑎 is equal to one and 𝑏 is equal to five, then we’re summing the first five terms in a series. We need to determine then what the values of 𝑎 and 𝑏 are for this series and find an expression for its general term.
Let’s have a look at the terms in this series then. We note, first of all, that all of the terms are fractions with a one in the numerator. And in the denominator, we have the sequence two, four, eight, 16, which are the powers of two. Two is two to the power of one. Four is two to the power of two. Eight is two to the power of three, and so on. So our general term is going to look something like one over two to the power of 𝑟 because when we substitute 𝑟 equals one, we get one over two. When we substitute 𝑟 equals two, we get one over two squared, which is four, and so on.
Notice though that the signs of these terms are alternating. The signs are negative for terms with an odd term number. And they’re positive for terms with an even term number. If we therefore include negative one to the power of 𝑟 in the numerator of our general term, this will give negative one for terms with an odd term number and positive one for terms with an even term number, which is what we want.
As both the numerator and denominator in this fraction are being raised to the same power of 𝑟, we can write the whole fraction to the power of 𝑟. So we have that our general term 𝑢𝑟 is negative one-half to the power of 𝑟.
Next, we need to determine the limits, the values of 𝑎 and 𝑏 for this sum. In our very first term, we were taking the value of 𝑟 to be one so that we got two to the power of one in the denominator of the fraction. Therefore, the value of 𝑎, the first value of 𝑟, must be one.
Notice that when the series was written in the question, there were these three dots after the fourth term, which indicates that the series continues. We’re not told where the series ends. So we assume that it continues indefinitely. We therefore have an infinite series. And so our upper limit for the sum is infinity.
Expressed in sigma notation then, we have that the series negative a half plus one-quarter minus one-eighth plus one sixteenth continuing is equal to the sum from 𝑟 equals one to infinity of negative one-half to the power of 𝑟.