Question

How do you use sigma notation to write the sum for $1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\cdots -\dfrac{1}{128}$?

Answer 1

Hint: We will look for a pattern in the terms given in the series. We will start with the denominators of the terms. Then we will take into account the fact that the sign is changing from plus to minus alternately. We will form a general term with a variable for the series. Then we will write this series using sigma notation.

Complete step by step answer:
The given series is $1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\cdots -\dfrac{1}{128}$. We will come up with a general term that will represent every term in this series. We will find a pattern in the terms to come up with the general term. We can see that the denominators are powers of 2. ${{2}^{0}}=1$; ${{2}^{1}}=2$; ${{2}^{3}}=8$, and so on upto ${{2}^{7}}=128$. Another thing we observe is that the sign changes alternately from a plus to a minus. We know that the even powers of $-1$ are 1 and the odd powers of $-1$ are $-1$. The first term can be written as $1=\dfrac{{{\left( -1 \right)}^{0}}}{{{2}^{0}}}$. The second term can be written as $-\dfrac{1}{2}=\dfrac{{{\left( -1 \right)}^{1}}}{{{2}^{1}}}$ and so on. So, now we can form a general term as \[{{\left( \dfrac{-1}{2} \right)}^{n}}\], where we start from $n=0$ and end at $n=7$. So, using the sigma notation, we can write the given series as
$1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\cdots -\dfrac{1}{128}={{\sum\limits_{n=0}^{7}{\left( \dfrac{-1}{2} \right)}}^{n}}$

Note:
The sigma notation is a compact way of writing the sum of a series of terms. Below the sigma symbol, we mention the starting value of the variable in the general term. Above the sigma symbol, we mention the ending value of the general term. We should be careful while forming the general term so that the sum of terms in a series is represented correctly.