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mixas84 [53]
3 years ago
10

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

Mathematics
2 answers:
Lilit [14]3 years ago
8 0

For starters,

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

Consider the nth 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}

Schach [20]3 years ago
4 0

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

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