Question

How do you find the fourth partial sum of `Sigma 5(1/2)^i` from i=1 to `oo`?

Answer 1

`75/16`

Explanation:

A series is an infinite sum, i.e. a sum with infinite terms.

Partial sums are approximations of the series, where you reduce it to a finite number of terms. In fact, the `n^"th"` partial sum is the sum of the first `n` terms, and the series is the limit of the partial sums as `n \to \infty`. You can imagine this, for example, as approaching the infinite sum by summing the first `10`, `100`, `1000`, `100000`, `1000000000`,... terms.

So, the fourth partial sum is the sum of the first `4` terms. Since `i` starts from one, we must sum the terms corresponding to `i=1,2,3,4`.

First of all, let's factor the `5` out of the sum:

`\sum_{i=1}^4 5 (1/2)^i = 5\sum_{i=1}^4 (1/2)^i`

Now we can simply translate the notations, plugging `1,2,3,4` in place of `i` for each term:

`5\sum_{i=1}^4 (1/2)^i = 5(1/2+1/4+1/8+1/16)5*15/16=75/16`