Question

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

plz help me​

Answer 1

3/16 (converges)

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

Answer 2

For starters,

`(3^k)/(4^(k+2))=(3^k)/(4^24^k)=\frac1{16}\left(\frac34\right)^k`

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

`S_n=\frac1{16}\left(\frac34\right)+\frac1{16}\left(\frac34\right)^2+\frac1{16}\left(\frac34\right)^3+\cdots+\frac1{16}\left(\frac34\right)^n`

Multiply both sides by
`\frac34`:

`\frac34S_n=\frac1{16}\left(\frac34\right)^2+\frac1{16}\left(\frac34\right)^3+\frac1{16}\left(\frac34\right)^4+\cdots+\frac1{16}\left(\frac34\right)^(n+1)`

Subtract
`S_n` from this:

`\frac34S_n-S_n=\frac1{16}\left(\frac34\right)^(n+1)-\frac1{16}\left(\frac34\right)`

Solve for
`S_n`:

`-\frac14S_n=\frac3{64}\left(\left(\frac34\right)^n-1\right)`

`S_n=\frac3{16}\left(1-\left(\frac34\right)^n\right)`

Now as
`n\to\infty`, the exponential term will converge to 0, since
`r^n\to0` if
`0<r<1`. This leaves us with

`\displaystyle\lim_(n\to\infty)S_n=\lim_(n\to\infty)\sum_(k=1)^n(3^k)/(4^(k+2))=\sum_(k=1)^\infty(3^k)/(4^(k+2))=\frac3{16}`