Question

Find the sum to infinite of the series $\dfrac{1}{3}+\dfrac{1}{{{5}^{2}}}+\dfrac{1}{{{3}^{3}}}+\dfrac{1}{{{5}^{4}}}+\dfrac{1}{{{3}^{5}}}+\dfrac{1}{{{5}^{6}}}+...$ .

Answer 1

To find the sum to infinite of the series $\dfrac{1}{3}+\dfrac{1}{{{5}^{2}}}+\dfrac{1}{{{3}^{3}}}+\dfrac{1}{{{5}^{4}}}+\dfrac{1}{{{3}^{5}}}+\dfrac{1}{{{5}^{6}}}+...$ , we have to rearrange the terms by combining the odd exponents of the denominator and the even exponents. Then, we can write the sum to infinity of the given series as the sum to infinity of the odd terms and the sum to infinity of the even terms. We have to find the sum to infinity of the odd terms and that of the even terms by using the formula for the sum to infinity a geometric series which is given by ${{S}_{\infty }}=\dfrac{a}{\left( 1-r \right)},r\ne 1$ since the terms in the even and odd series forms a geometric series. a in the formula is the first term and r is the common ratio which is obtained by dividing the second term by the first.

Complete step by step answer:
We have to find the sum to infinite of the series $\dfrac{1}{3}+\dfrac{1}{{{5}^{2}}}+\dfrac{1}{{{3}^{3}}}+\dfrac{1}{{{5}^{4}}}+\dfrac{1}{{{3}^{5}}}+\dfrac{1}{{{5}^{6}}}+...$ . We can rewrite the given series by combining the odd exponents of the denominator and the even exponents.
$\Rightarrow \underbrace{\dfrac{1}{3}+\dfrac{1}{{{3}^{3}}}+\dfrac{1}{{{3}^{5}}}+...}_{odd}+\underbrace{\dfrac{1}{{{5}^{2}}}+\dfrac{1}{{{5}^{4}}}+\dfrac{1}{{{5}^{6}}}+...}_{even}$
Therefore, we can write the sum to infinity of the given series as the sum to infinity of the odd terms and the sum to infinity of the even terms.
${{S}_{\infty }}={{S}_{odd}}+{{S}_{even}}...\left( i \right)$
Let us find the ${{S}_{odd}}$ , that is sum of the series $\dfrac{1}{3}+\dfrac{1}{{{3}^{3}}}+\dfrac{1}{{{3}^{5}}}+...$ . We can see that the given series is of the form $a+ar+a{{r}^{2}}+...$ which is a geometric series with first term, ${{a}_{o}}$ as $\dfrac{1}{3}$ . Let us find the common ratio by dividing the second term by the first.
$\begin{aligned} & \Rightarrow {{r}_{0}}=\dfrac{\dfrac{1}{{{3}^{3}}}}{\dfrac{1}{3}} \\\ & \Rightarrow {{r}_{0}}=\dfrac{1}{{{3}^{2}}} \\\ \end{aligned}$

We know that the sum to infinity a geometric series is given by
${{S}_{\infty }}=\dfrac{a}{\left( 1-r \right)},r\ne 1...\left( ii \right)$
where a is the first term of the series and r is the common ratio.
Therefore, the sum to infinity of the odd terms can be found by substituting the values in the formula (ii).
$\begin{aligned} & \Rightarrow {{S}_{odd}}=\dfrac{\dfrac{1}{3}}{\left( 1-\dfrac{1}{{{3}^{2}}} \right)} \\\ & \Rightarrow {{S}_{odd}}=\dfrac{\dfrac{1}{3}}{\left( \dfrac{9-1}{9} \right)} \\\ & \Rightarrow {{S}_{odd}}=\dfrac{\dfrac{1}{3}}{\dfrac{8}{9}} \\\ & \Rightarrow {{S}_{odd}}=\dfrac{1}{\require{cancel}\cancel{3}}\times \dfrac{{{\require{cancel}\cancel{9}}^{3}}}{8} \\\ & \Rightarrow {{S}_{odd}}=\dfrac{3}{8}...\left( iii \right) \\\ \end{aligned}$
Let us find the sum to infinity of the even terms. Firstly, we have to find the first term and the common ratio of the even series $\dfrac{1}{{{5}^{2}}}+\dfrac{1}{{{5}^{4}}}+\dfrac{1}{{{5}^{6}}}+...$ .
$\begin{aligned} & \Rightarrow {{r}_{e}}=\dfrac{\dfrac{1}{{{5}^{4}}}}{\dfrac{1}{{{5}^{2}}}} \\\ & \Rightarrow {{r}_{e}}=\dfrac{1}{{{5}^{2}}} \\\ \end{aligned}$
The first term is ${{a}_{e}}=\dfrac{1}{{{5}^{2}}}$ .
Therefore, sum to infinity of the even terms is given by
$\begin{aligned} & \Rightarrow {{S}_{even}}=\dfrac{\dfrac{1}{{{5}^{2}}}}{\left( 1-\dfrac{1}{{{5}^{2}}} \right)} \\\ & \Rightarrow {{S}_{even}}=\dfrac{\dfrac{1}{{{5}^{2}}}}{\dfrac{25-1}{25}} \\\ & \Rightarrow {{S}_{even}}=\dfrac{1}{24}...\left( iv \right) \\\ \end{aligned}$
Let us substitute (iii) and (iv) in (i).
$\begin{aligned} & \Rightarrow {{S}_{\infty }}=\dfrac{3}{8}+\dfrac{1}{24} \\\ & \Rightarrow {{S}_{\infty }}=\dfrac{9+1}{24} \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow {{S}_{\infty }}=\dfrac{10}{24} \\\ & \Rightarrow {{S}_{\infty }}=\dfrac{5}{12} \\\ \end{aligned}$
Therefore, the sum to infinite of the series $\dfrac{1}{3}+\dfrac{1}{{{5}^{2}}}+\dfrac{1}{{{3}^{3}}}+\dfrac{1}{{{5}^{4}}}+\dfrac{1}{{{3}^{5}}}+\dfrac{1}{{{5}^{6}}}+...$ is $\dfrac{5}{12}$ .

Note: Students must be thorough with standard forms of arithmetic and geometric series. They must deeply learn the formulas of these series. Students must note that the formula for the sum to infinity of a GP and the sum of n terms of the GP are different. The formula for the latter is ${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)},r\ne 1$ .