Question

# Calculate the sum of the following sequence:0.45, 0.015, 0.0005,â€¦

Hint: Try to find a pattern. What kind of series is it? A geometric series, but take care that the given series is infinite, so donâ€™t use the finite series formula. So use the sum of terms of the infinite series formula.

According to the question, we have to find the value of S, i.e.,
S = 0.45 + 0.015 + 0.005â€¦â€¦â€¦..
This is an infinite sum.
Let, $S={{a}_{1}}+{{a}_{2}}+{{a}_{3}}........$
Where, ${{a}_{1}}=0.45,{{a}_{2}}=0.015,{{a}_{3}}=0.005$ and so on
Now, divide the first two term, we get
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{0.45}{0.015}=30$
Now divide the second and the third term, we get
$\dfrac{{{a}_{2}}}{{{a}_{3}}}=\dfrac{0.015}{0.005}=30$
Here we can observe that the series has a common ratio, that is $r=\dfrac{1}{30}$ .
Hence the given series is infinite geometric progression.
So the given infinite series can be written in the form of,
$S=a+ar+a{{r}^{2}}.......$
Where â€˜aâ€™ is the first term and â€˜râ€™ is the common ratio, i.e,
$S=0.45+\dfrac{0.45}{\left( 30 \right)}+\dfrac{0.45}{{{\left( 30 \right)}^{2}}}........$
We know the sum of infinite geometric progression is given by the formula,
$S=\dfrac{a}{1-r}$
Substituting the values of â€˜aâ€™ and â€˜râ€™, we get
$S=\dfrac{0.45}{1-\dfrac{1}{30}}$
Taking the LCM in the denominator, we get
\begin{align} & S=\dfrac{0.45}{\dfrac{30-1}{30}} \\ & \Rightarrow S=\dfrac{0.45}{\dfrac{29}{30}} \\ \end{align}
We know division is also written as the multiplication of the inverse of the number, so the above equation can be written as,
\begin{align} & S=0.45\times \dfrac{30}{29} \\ & \Rightarrow S=0.45\times \dfrac{30}{29} \\ & S=\dfrac{13.5}{29}=0.4655 \\ \end{align}
So the sum of 0.45, 0.015, 0.0005,â€¦ is 0.4655.

Note: There is no need to get confused by decimals. Also, students often think the infinite sum is infinite. This is not true, students also take $\dfrac{{{a}_{1}}}{{{a}_{2}}}$ to be r, when $r=\dfrac{{{a}_{2}}}{{{a}_{1}}}.$
Sometimes student get confused with the infinite series and apply the formula of finite geometric progression formula, i.e.
$S=\dfrac{a(1-{{r}^{n}})}{1-r}$
This will give the wrong answer.