Question

Can someone help me find the sum? Not sure how to start. All I know is that it is convergent.

Answer 1

there is a formula for sum of a 'geometric series'

Answer 2

Okay, thank. I know it kinda but I'm just not really sure how to plug everything in. And I don't exactly understand what I'm trying to find the sum of, exactly.

Answer 3

*thanks

Answer 4

I only know how to find a partial sum

Answer 5

$$
\Large{

\sum_{i=1}^{\infty } 8~ (\frac56)^{i-1}
\\= 8~ (\frac56)^{1-1} + 8~ (\frac56)^{2-1} + 8~ (\frac56)^{3-1} + ...
\\=\\= 8~ (\frac56)^{0} + 8~ (\frac56)^{1} + 8~ (\frac56)^{2} + ...

}
$$

Answer 6

ok do you see that the sum is a shorthand notation for a lot of summands

Answer 7

we want to know if that sum converges

Answer 8

Ohhhhh okay this is starting to look familiar

Answer 9

$$
\Large{

\sum_{i=1}^{\infty } 8~ (\frac56)^{i-1}
\\= 8~ (\frac56)^{1-1} + 8~ (\frac56)^{2-1} + 8~ (\frac56)^{3-1} + ...
\\\\= 8~ (\frac56)^{0} + 8~ (\frac56)^{1} + 8~ (\frac56)^{2} + ...
\\ =8~ [ (\frac56)^{0} + (\frac56)^{1} + (\frac56)^{2} + ...
\\ =8~ ( 1 + \frac56 + \frac{25}{36} + ... )
}
$$

Answer 10

you might have seen that
$$
\Large 1+ \frac12 + \frac 14 + \frac 18 + ... = 2
$$

Answer 11

but not all series which have smaller terms converges. for example:
$$
\Large 1 + \frac12 +\frac 13 + \frac 14 + ... = \infty

$$

Answer 12

It can be shown that any series which can be written as a 'geometric' series converges.
as long as r is less than 1.

$$
\Large{


\sum_{i=1}^{\infty } a\cdot r^{i-1} = \frac{a}{1-r}
\\ \text{ when |r| <1 }
}
$$

Answer 13

this type of series is called geometric , notice that you have a constant raised to a power, but the constant is less than 1

Answer 14

you could solve this as well by looking at the partial sums, would you like to do that?

Answer 15

No, its fine, I know how to do partial sums. I was just a bit confused with this question because there was no 'partial' but I understand now so thanks a bunch!

Answer 16

the partial sums are adding a finite number of terms
$$
\Large{

\\S_1= 8~ (\frac56)^{1-1}
\\S_2= 8~ (\frac56)^{1-1} + 8~ (\frac56)^{2-1}
\\ S_3= 8~ (\frac56)^{1-1} + 8~ (\frac56)^{2-1} + 8~ (\frac56)^{3-1}

}

$$

Answer 17

Wait, I have a question about the first equation I had

Answer 18

Is it possible for it to have a sum of 48?

Answer 19

Or would you say that it has no sum or a sum cannot be found?

Answer 20

it has a sum of exactly 64/3

Answer 21

maybe i can look at your work and see how you got 48