Hybrid Progression

Calculus Level 3

$\begin{aligned} && 3, 6,9,12,\ldots \\ && 3, 9,27,81,\ldots \\ \end{aligned}$

The above shows two rows of numbers, each of these rows has infinitely many numbers.
The first row of numbers follows an arithmetic progression , whereas
The second row of numbers follows a geometric progression .

We know that the sum of all the numbers in each row diverges to infinity .
However, the sum of the ratio of each term exists! What is it?

In other words, what is the value of

$\dfrac33 + \dfrac69 + \dfrac9{27} + \dfrac{12}{81} + \cdots \, ?$

The answer is 2.25.

2 solutions

Sabhrant Sachan
Jun 25, 2016

$\quad n^{th} \text{ Term of sequence : } T_{n} = \dfrac{3n}{3^n} \implies \dfrac{n}{3^{n-1}} \quad , \quad n \in \mathbb N \\\quad \text{Let the sequence be S} \\\quad S = \dfrac{1}{3^0}+ \dfrac{2}{3^1}+ \dfrac{3}{3^2}+\cdots+ \dfrac{n}{3^{n-1} } \\ \quad -\dfrac{1}{3}S =-\dfrac{1}{3^1}- \dfrac{2}{3^2}-\cdots-\dfrac{n-1}{3^{n-1} }-\dfrac{n}{3^{n} } \\ \quad \text{Add these two equation } \\ \quad \dfrac{2}{3}S= 1+\dfrac13+\dfrac{1}{3^2}+\cdot+\dfrac{1}{3^{n-1}}-\dfrac{n}{3^{n}} \\ \quad \dfrac{2}{3}S = \dfrac{1-\frac{1}{3^n}}{1-\frac13}-\dfrac{n}{3^n} \\ \quad \boxed{S=\dfrac{3}{2}\left(\dfrac{3}{2}\left(1-\frac{1}{3^n}\right)-\dfrac{n}{3^n}\right)} \\ \quad \displaystyle \lim_{n \to \infty} S = \dfrac{9}{4} = \boxed{2.25}$

Rajbir Malik
Jun 25, 2016

n/(3^(n-1))=(1/3)×(n-1)/(3^(n-2))+1/(3^(n-1)) Taking summation from n=1 to infinity, we get, S=S/3 + 1.5, => S=2.25

Hybrid Progression

The answer is 2.25.

2 solutions

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