Question

Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.

Answer 1

For starters,

$\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k$

Consider the $n$th partial sum, denoted by $S_n$:

$S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n$

Multiply both sides by $\frac34$:

$\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}$

Subtract $S_n$ from this:

$\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)$

Solve for $S_n$:

$-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)$

$S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)$

Now as $n\to\infty$, the exponential term will converge to 0, since $r^n\to0$ if $0$. This leaves us with

$\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}$

Answer 2

Answer:

3/16 (converges)

Step-by-step explanation:

Let's write out the first few terms of this sequence, from k=1 to k=3. This gives us:

sum = (3^1)/(4^(1+2)) + (3^2)/(4^(2+2)) + (3^3)/(4^(3+2)) + ...

Computing that into numbers, we have:

sum = 3/64 + 9/256 + 27/1024 + ...

Now, what happens if we multiply both sides by 4/3 (which we get from the 3 and the 4 in the problem)? This gives us:

(4/3)*sum = 4/3*(3/64) + 4/3*(9/256) + 4/3*(27/1024) + ...

which computes out to:

(4/3)*sum = 1/16 + 3/64 + 9/256 + ...

Now, if we subtract sum from 4/3*(sum), notice that most of the terms cancel out. We are left with:

(4/3)*sum - sum = 1/16

Solving this algebraic equation gives us sum = 3/16