$\displaystyle (1/1) + (4/4) + (7/16) + (10/64) ...$
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?
$\displaystyle (1/1) + (4/4) + (7/16) + (10/64) ...$
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?
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: .$\displaystyle S \:=\:\frac{8}{3}$
**
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:
$\displaystyle S=\lim_{n\to\infty}S_n=\frac{8}{3}$