Question

Let we have a series as \[S=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...\] . Find the sum of infinite terms of the series

Answer 1

Hint: We first check whether the series is geometric or not. After that, we write the formula for the summation of a geometric series which is $S=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ . Noting that $n\to \infty ,r<1$ , we can rewrite it as $S=\dfrac{a}{1-r}$ . Putting the values $a=1,r=\dfrac{1}{2}$ , we get the answer.

Complete step-by-step solution:
We need to check what type of series it is. For this, we need to find the relation between the consecutive terms. Let us check the difference between the consecutive terms. $1-\dfrac{1}{2}=\dfrac{1}{4}$ , \[\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\] , \[\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{1}{8}\] . We can see that the difference between the consecutive terms is not constant or common. So, we can say that there is no common difference and thus this is not an arithmetic series. Next, we need to check if it is a geometric series. Let us check the ratio between the consecutive terms. \[\dfrac{1}{2}\div 1=\dfrac{1}{2}\] , \[\dfrac{1}{4}\div \dfrac{1}{2}=\dfrac{1}{2}\] , \[\dfrac{1}{8}\div \dfrac{1}{4}=\dfrac{1}{2}\] . We can see that the ratio between the consecutive terms is a constant. So, we can say that the common ratio between the consecutive terms is \[\dfrac{1}{2}\] and it is a geometric series. The sum of a geometric series is $S=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ where, a is the first term, r is the common ratio and n is the number of terms. Since the number of terms is infinite and the common ratio is less than $1$ , so ${{r}^{n}}\to 0$ . The sum becomes $S=\dfrac{a}{1-r}$ . Putting the values, we get,
\[S=\dfrac{1}{1-\dfrac{1}{2}}=2\]
Thus, we can conclude that the sum is $2$ .

Note: We must be careful enough to use the formula $S=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ instead of the formula $S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ since here r is less than $1$ . Also, we should note that the common ratio is $\dfrac{1}{2}$ and not $2$ .