# Thread: Infinite Series with Numerator Arithmetic and Denominator Geometric

1. ## Infinite Series with Numerator Arithmetic and Denominator Geometric

$\displaystyle (1/1) + (4/4) + (7/16) + (10/64) ...$

This series is supposed to be an infinite series but it's not going up by a set common ratio. How do I find the sum of the infinite series of this series?

2. ## Re: Infinite Series with Numerator Arithmetic and Denominator Geometric

Hello, Espionage!

$\displaystyle \text{Evaluate: }\:S \;=\;\frac{1}{1} + \frac{4}{4} + \frac{7}{4^2} + \frac{10}{4^3} + \cdots$

$\displaystyle \begin{array}{ccccccc}\text{We are given:}&S &=& \dfrac{1}{1} + \dfrac{4}{4} + \dfrac{7}{4^2} + \dfrac{10}{4^3} + \cdots \\ \\[-3mm] \text{Multiply by }\frac{1}{4}\!: & \frac{1}{4}S &=& \quad\;\;\; \dfrac{1}{4} + \dfrac{4}{4^2} + \dfrac{7}{4^3} + \cdots \\ \\[-3mm] \text{Subtract:} & \frac{3}{4}S &=& 1 + \underbrace{\frac{3}{4} + \frac{3}{4^2} + \frac{3}{4^3} + \cdots}_{\text{geometric series}} \end{array}$

The geometric series has first term $\displaystyle a = \tfrac{3}{4}$ and common ratio $\displaystyle r = \tfrac{1}{4}$

Its sum is: .$\displaystyle \dfrac{\frac{3}{4}}{1-\frac{1}{4}} \:=\:\dfrac{\frac{3}{4}}{\frac{3}{4}} \:=\:1$

Hence: .$\displaystyle \tfrac{3}{4}S \:=\:1+1 \:=\:2$

Therefore: .$\displaystyle S \:=\:\frac{8}{3}$

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3. ## Re: Infinite Series with Numerator Arithmetic and Denominator Geometric

Edit: Post above answered it way better.

4. ## Re: Infinite Series with Numerator Arithmetic and Denominator Geometric

If I was going to find the infinite sum, I would begin with the difference equation to obtain the partial sum in closed form:

$\displaystyle S_{n}-S_{n-1}=\frac{3n+1}{4^n}$ where $\displaystyle S_0=1$

The corresponding homogeneous solution is:

$\displaystyle h_n=c_1$

We then look for a particular solution of the form:

$\displaystyle p_n=(An+B)4^{-n}$

We may now determine the coefficients A and B.

Substituting this into the original difference equation, we find:

$\displaystyle (An+B)4^{-n}-(A(n-1)+B)4^{-(n-1)}=(3n+1)4^{-n}$

Multiply through by $\displaystyle 4^{n}$:

$\displaystyle (An+B)-4(A(n-1)+B)=3n+1$

$\displaystyle -3An+4A-3B=3n+1$

Equating coefficients, we obtain the system:

$\displaystyle -3A=3\,\therefore\,A=-1$

$\displaystyle 4A-3B=1\,\therefore\,B=-\frac{5}{3}$

And thus:

$\displaystyle p_n=-\frac{1}{3}(3n+5)4^{-n}$

By superposition, we have:

$\displaystyle S_n=h_n+p_n=c_1-\frac{1}{3}(3n+5)4^{-n}$

Using the initial value, we find:

$\displaystyle S_0=c_1-\frac{1}{3}(3(0)+5)4^{0}=c_1-\frac{5}{3}=1\,\therefore\,c_1=\frac{8}{3}$

and so we find:

$\displaystyle S_n=\frac{8}{3}-\frac{1}{3}(3n+5)4^{-n}$

Hence, the infinite sum is:

$\displaystyle S=\lim_{n\to\infty}S_n=\frac{8}{3}$

5. ## Re: Infinite Series with Numerator Arithmetic and Denominator Geometric

Thank you very much.

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# numerator is in AP and denominator is in GP

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