Question

Calculate the sum of the series $3,\sqrt{3},1,....$

Answer 1

Hint: Consider the sequence and find out the common ratio and find out whether it is greater than or less than one. But be careful of the series, it is an infinite series and not a finite series.

In the question we have been given a task to find the sum of $3,\sqrt{3},1,....$
The sequence given is actually in GP, that is, a geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non – zero number called common ratio.
So, in $3,\sqrt{3},1,....$ the common ratio can be found out by dividing the second term by ${{1}^{st}}$ term which is $\dfrac{\sqrt{3}}{3}=\dfrac{1}{\sqrt{3}}$.
Now, by observing we see that the common ratio is equal to $\dfrac{1}{\sqrt{3}}$ this is less than one.
So we know that the sum of infinite series in GP is given by,
$sum=\dfrac{a}{1-r}$
Where ‘a’ is the first term and ‘r’ is the common ratio.
In our question, a = 3 which signifies the ${{1}^{st}}$term and $r=\dfrac{1}{\sqrt{3}}$ which signify the common ratio.
So, putting the values a = 3 and $r=\dfrac{1}{\sqrt{3}}$ we get;
$sum=\dfrac{3}{1-\dfrac{1}{\sqrt{3}}}=\dfrac{3}{\dfrac{\sqrt{3}-1}{\sqrt{3}}}=\dfrac{3\sqrt{3}}{\sqrt{3}-1}$
Now, multiplying the conjugate to numerator and denominator we get;
$\begin{align}
  & sum=\dfrac{3\sqrt{3}}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1} \\
 & =\dfrac{3\sqrt{3}\left( \sqrt{3}+1 \right)}{3-1}=\dfrac{3\sqrt{3}}{2}\left( \sqrt{3}+1 \right) \\
\end{align}$
Hence, the sum of the given series is $\dfrac{3\sqrt{3}}{2}\left( \sqrt{3}+1 \right)$

Note: In the sequence of G.P, whose common ratio is less than one; their value converges that is why their infinite sum is a finite number. While in the sequences of $G.P.$whose common ratio is greater than one, their value diverges that is why their infinite sum is an infinite number.
Students generally make mistakes by taking the sum of finite GP series instead of infinite GP series. This will give the wrong answer.