# Infinite Series with Numerator Arithmetic and Denominator Geometric

• Jul 12th 2013, 06:31 PM
Espionage
Infinite Series with Numerator Arithmetic and Denominator Geometric
$(1/1) + (4/4) + (7/16) + (10/64) ...$

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?
• Jul 12th 2013, 07:01 PM
Soroban
Re: Infinite Series with Numerator Arithmetic and Denominator Geometric
Hello, Espionage!

Quote:

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

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

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

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

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

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

**
• Jul 12th 2013, 07:04 PM
chiro
Re: Infinite Series with Numerator Arithmetic and Denominator Geometric
Edit: Post above answered it way better.
• Jul 12th 2013, 07:24 PM
MarkFL
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:

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

The corresponding homogeneous solution is:

$h_n=c_1$

We then look for a particular solution of the form:

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

We may now determine the coefficients A and B.

Substituting this into the original difference equation, we find:

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

Multiply through by $4^{n}$:

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

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

Equating coefficients, we obtain the system:

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

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

And thus:

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

By superposition, we have:

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

Using the initial value, we find:

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

and so we find:

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

Hence, the infinite sum is:

$S=\lim_{n\to\infty}S_n=\frac{8}{3}$
• Jul 12th 2013, 07:26 PM
Espionage
Re: Infinite Series with Numerator Arithmetic and Denominator Geometric
Thank you very much.