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Question: How do you know when a geometric series converges?...

How do you know when a geometric series converges?

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Solution

Here in this question we have been asked to explain how we can say that a geometric series converges. The actual and simple meaning of a series being convergent is under some certain conditions the sum of the series is a finite value for an infinite number of terms. If not then the series is said to be divergent.

Complete step by step solution:
Now considering from the question we have been asked to explain how we can say that a geometric series converges.
From the basic concepts we know that the terms in geometric series has a common ratio which can be expressed as ∑arn−1\sum{a{{r}^{n-1}}} where rr is the common ratio and aa is the first term. The sum of the series is given as Sn=a(1−rn)1−r{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r} .
When n→∞n \to \infty the value of the sum of the series is generally given as lim⁡n→∞Sn=lim⁡n→∞(a(1−rn)1−r) ⇒lim⁡n→∞(a1−r−arn1−r)=lim⁡n→∞(a1−r)−lim⁡n→∞((arn)1−r) ⇒a1−r−(a1−r)lim⁡n→∞rn=a1−r \begin{aligned} & \displaystyle \lim_{n \to \infty }{{S}_{n}}=\displaystyle \lim_{n \to \infty }\left( \dfrac{a\left( 1-{{r}^{n}} \right)}{1-r} \right) \\\ & \Rightarrow \displaystyle \lim_{n \to \infty }\left( \dfrac{a}{1-r}-\dfrac{a{{r}^{n}}}{1-r} \right)=\displaystyle \lim_{n \to \infty }\left( \dfrac{a}{1-r} \right)-\displaystyle \lim_{n \to \infty }\left( \dfrac{\left( a{{r}^{n}} \right)}{1-r} \right) \\\ & \Rightarrow \dfrac{a}{1-r}-\left( \dfrac{a}{1-r} \right)\displaystyle \lim_{n \to \infty }{{r}^{n}}=\dfrac{a}{1-r} \\\ \end{aligned}
The actual and simple meaning of a series being convergent is under some certain conditions the sum of the series is a finite value for an infinite number of terms. If not then the series is said to be divergent.
Generally a geometric series converges when −1<r<1-1 < r < 1 that is ∣r∣<1\left| r \right| < 1 and its value is given as a1−r\dfrac{a}{1-r} .
Therefore we can conclude that a geometric series converges when −1<r<1-1 < r < 1 that is ∣r∣<1\left| r \right| < 1 and its value is given as a1−r\dfrac{a}{1-r}.

Note: While answering questions of this type we should be sure with the concepts that we are going to apply in the process. Similarly we can verify the convergence for any other series. Someone can make a mistake if they are not clear with their concepts that in case if you have taken the sum of the geometric series as Sn=arn1−r{{S}_{n}}=\dfrac{a{{r}^{n}}}{1-r} which a wrong formula we will be ending up having a wrong conclusion.