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Question: How do you find the sum of the infinite geometric series \[1-\dfrac{1}{5}+\dfrac{1}{25}-\dfrac{1}{12...

How do you find the sum of the infinite geometric series 1−15+125−1125+...1-\dfrac{1}{5}+\dfrac{1}{25}-\dfrac{1}{125}+...?

Explanation

Solution

The sum of infinite terms of a geometric progression with a common ratio less than 1 is a1−r\dfrac{a}{1-r}. Here, a is the first term of the geometric series, and r is the common ratio of the series. We can find the common ratio by taking the ratio of a term with its previous term. To find the sum of the series we first have to find the first term and common ratio, and then substitute it in the summation formula.

Complete step-by-step answer:
We are given the infinite geometric series 1−15+125−1125+...1-\dfrac{1}{5}+\dfrac{1}{25}-\dfrac{1}{125}+.... Here, the first term is 1, so a=1a=1. To find the common ratio, we need to take a ratio of a term with its previous term. Hence, we get the ratio as
r=−151=−15r=\dfrac{-\dfrac{1}{5}}{1}=-\dfrac{1}{5}
Now, we have the first term and the common ratio. Substituting their values in the formula for the sum of an infinite geometric series. We get
⇒a1−r=11−(−15)=11+15\Rightarrow \dfrac{a}{1-r}=\dfrac{1}{1-\left( -\dfrac{1}{5} \right)}=\dfrac{1}{1+\dfrac{1}{5}}
Simplifying the above calculation, we get
⇒165\Rightarrow \dfrac{1}{\dfrac{6}{5}}
Multiplying the above fraction by 55\dfrac{5}{5}, we get
⇒165×55=56\Rightarrow \dfrac{1}{\dfrac{6}{5}}\times \dfrac{5}{5}=\dfrac{5}{6}
Hence, the sum of the given infinite geometric series is 56\dfrac{5}{6}.

Note: For a general geometric series the formula for the sum of n terms is, a(1−rn)1−r\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r} for ∣r∣<1\left| r \right|<1, and a(rn−1)r−1\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1} for ∣r∣>1\left| r \right|>1. We can find the sum of infinite series only if the absolute value of the common ratio is less than one, that is ∣r∣<1\left| r \right|<1.
We can derive the formula for infinite geometric series as,
lim⁡n→∞a(1−rn)1−r\displaystyle \lim_{n \to \infty }\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}
As ∣r∣<1\left| r \right|<1, we can say that . Using this in the above limit, we get the summation formula as
lim⁡n→∞a(1−rn)1−r=a(1−0)1−r=a1−r\displaystyle \lim_{n \to \infty }\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}=\dfrac{a\left( 1-0 \right)}{1-r}=\dfrac{a}{1-r}