Question

Find the sum of the infinite series: \[\dfrac{1.2}{3}+\dfrac{2.3}{{{3}^{2}}}+\dfrac{3.4}{{{3}^{3}}}+\dfrac{4.5}{{{3}^{4}}}+.....\]

Answer 1

Hint: The given series doesn’t fall in any kind of known sum of series. So, we need to change its form by using multiplication. We slide 1 number and subtract them. We keep continuing this process until we get a solution of known formula. As the sum is infinite, all the elements will follow the rules applied.

Complete answer:
We try to find out the general term of the equation from looking at the form.
So, ${{t}_{n}}$ be the nth term of the sequence.
Then ${{t}_{1}}=\dfrac{1.2}{{{3}^{1}}}=\dfrac{1.\left( 1+1 \right)}{{{3}^{1}}}$, ${{t}_{2}}=\dfrac{2.3}{{{3}^{2}}}=\dfrac{2.\left( 2+1 \right)}{{{3}^{2}}}$, ${{t}_{3}}=\dfrac{3.4}{{{3}^{3}}}=\dfrac{3.\left( 3+1 \right)}{{{3}^{3}}}$, ${{t}_{4}}=\dfrac{4.5}{{{3}^{4}}}=\dfrac{4.\left( 4+1 \right)}{{{3}^{4}}}$.
So, the general term will be ${{t}_{n}}=\dfrac{n.\left( n+1 \right)}{{{3}^{n}}}$.
So, the sum of the series be ${{S}_{n}}=\sum\limits_{n=1}^{\infty }{{{t}_{n}}}$. We multiply both sides with $\dfrac{1}{3}$ to get $\dfrac{1}{3}{{S}_{n}}=\sum\limits_{n=1}^{\infty }{\dfrac{{{t}_{n}}}{3}}$.
We have two equations ${{S}_{n}}=\sum\limits_{n=1}^{\infty }{{{t}_{n}}}=\dfrac{1.2}{3}+\dfrac{2.3}{{{3}^{2}}}+\dfrac{3.4}{{{3}^{3}}}+\dfrac{4.5}{{{3}^{4}}}+.....$ and $\dfrac{1}{3}{{S}_{n}}=\sum\limits_{n=1}^{\infty }{\dfrac{{{t}_{n}}}{3}}=\dfrac{1.2}{{{3}^{2}}}+\dfrac{2.3}{{{3}^{3}}}+\dfrac{3.4}{{{3}^{4}}}+\dfrac{4.5}{{{3}^{5}}}+.....$.
We subtract second equation from the first one and follow the denominator of the individual elements to subtract them to get a new series
\[\begin{align}
  & {{S}_{n}}-\dfrac{1}{3}{{S}_{n}}=\left( \dfrac{1.2}{3}+\dfrac{2.3}{{{3}^{2}}}+\dfrac{3.4}{{{3}^{3}}}+\dfrac{4.5}{{{3}^{4}}}+..... \right)-\left( \dfrac{1.2}{{{3}^{2}}}+\dfrac{2.3}{{{3}^{3}}}+\dfrac{3.4}{{{3}^{4}}}+\dfrac{4.5}{{{3}^{5}}}+..... \right) \\
 & \Rightarrow \dfrac{2}{3}{{S}_{n}}=\left[ \dfrac{1.2}{3}+\left( \dfrac{2.3}{{{3}^{2}}}-\dfrac{1.2}{{{3}^{2}}} \right)+\left( \dfrac{3.4}{{{3}^{3}}}-\dfrac{2.3}{{{3}^{3}}} \right)+\left( \dfrac{4.5}{{{3}^{4}}}-\dfrac{3.4}{{{3}^{4}}} \right)+..... \right] \\
 & \Rightarrow \dfrac{2}{3}{{S}_{n}}=\left[ \dfrac{1.2}{3}+\dfrac{2.2}{{{3}^{2}}}+\dfrac{3.2}{{{3}^{3}}}+\dfrac{4.2}{{{3}^{4}}}+..... \right] \\
\end{align}\]
We get a new sum of infinite series \[{{S}_{n}}=\left[ 1+\dfrac{2}{3}+\dfrac{3}{{{3}^{2}}}+\dfrac{4}{{{3}^{3}}}+..... \right]\].
The whole previous process, we apply 1 more time. We multiply the series with $\dfrac{1}{3}$.
We get \[\dfrac{1}{3}{{S}_{n}}=\left[ \dfrac{2}{3}+\dfrac{2}{{{3}^{2}}}+\dfrac{3}{{{3}^{3}}}+\dfrac{4}{{{3}^{4}}}+..... \right]\]
We subtract second equation from the first one and follow the denominator of the individual elements to subtract them.
\[\begin{align}
  & {{S}_{n}}-\dfrac{1}{3}{{S}_{n}}=\left[ 1+\dfrac{2}{3}+\dfrac{3}{{{3}^{2}}}+\dfrac{4}{{{3}^{3}}}+..... \right]-\left[ \dfrac{2}{3}+\dfrac{2}{{{3}^{2}}}+\dfrac{3}{{{3}^{3}}}+\dfrac{4}{{{3}^{4}}}+..... \right] \\
 & \Rightarrow \dfrac{1}{3}{{S}_{n}}=\dfrac{2}{3}+\left( 1-\dfrac{2}{3} \right)+\left( \dfrac{3}{{{3}^{2}}}-\dfrac{2}{{{3}^{2}}} \right)+\left( \dfrac{4}{{{3}^{3}}}-\dfrac{3}{{{3}^{3}}} \right)+..... \\
 & \Rightarrow \dfrac{1}{3}{{S}_{n}}=\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{3}}}+..... \\
 & \Rightarrow \dfrac{{{S}_{n}}}{3}=1+\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{3}}}+..... \\
 & \Rightarrow \dfrac{{{S}_{n}}}{3}+\dfrac{1}{3}=1+\dfrac{1}{3}+\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{3}}}+..... \\
\end{align}\]
So, the right side of the equation becomes a sum of infinite G.P. whose common ratio is less than 1.
So, \[\dfrac{{{S}_{n}}+1}{3}=1+\dfrac{1}{3}+\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{3}}}+.....=\dfrac{1}{1-\dfrac{1}{3}}=\dfrac{3}{2}\]. The formula for infinite G.P. is $\dfrac{a}{1-r}$ where a = first element and r is the common ratio.
Now we get the solution of the series as \[\dfrac{{{S}_{n}}+1}{3}=\dfrac{3}{2}\Rightarrow {{S}_{n}}=\dfrac{3\times 3}{2}-1=\dfrac{7}{2}\].
So, the sum of the infinite series: \[\dfrac{1.2}{3}+\dfrac{2.3}{{{3}^{2}}}+\dfrac{3.4}{{{3}^{3}}}+\dfrac{4.5}{{{3}^{4}}}+.....\] is $\dfrac{7}{2}$.

Note:
 We don’t need to go to all the end of the series. The starting number pattern will tell us the general terms of the series. Also, need to be careful about the subtraction method we are applying here. The common ration r in the final G.P. series has to be in the form $\left| r \right|<1$. So, we need to be careful about it.