Question

Consider the infinite geometric series ∑∞ n=1 -4(1/3)^n-1

Answer 1

a. The series is

$\displaystyle\sum_{n=1}^\infty-4\left(\frac13\right)^{n-1}=-4-\frac43-\frac4{3^2}-\frac4{3^3}-\cdots$

(first four terms are listed)

b. The series converges because this is a geometric series with $r=\dfrac13$.

c. Let $S_N$ be the $N$-th partial sum of the series:

$S_N=\displaystyle\sum_{n=1}^N-4\left(\frac13\right)^{n-1}$

$S_N=-4-\dfrac43-\dfrac4{3^2}-\cdots-\dfrac4{3^{N-1}}$

Multiplying both sides by $\dfrac13$ gives

$\dfrac13S_N=-\dfrac43-\dfrac4{3^2}-\dfrac4{3^3}-\cdots-\dfrac4{3^N}$

Subtracting this from $S_N$ gives

$S_N-\dfrac13S_N=\dfrac23S_N=-4+\dfrac4{3^N}$

$\implies S_N=-6+\dfrac6{3^N}$

As $N$ gets larger and larger $(N\to\infty)$ the rational term converges to 0 and we're left with

$\displaystyle\lim_{N\to\infty}S_N=\sum_{n=1}^\infty-4\left(\frac13\right)^{n-1}=-6$