How do you determine if #a_n=6+19+3+4/25+8/125+16/625+...+(2/5)^n+...# converge and find the sums when they exist?

From the fourth term on we have a geometric series.

The sum of this can be found using:

So from 4th term to infinity we have:

Then we have:

And:

To determine if the series (a_n = 6 + 19 + 3 + \frac{4}{25} + \frac{8}{125} + \frac{16}{625} + \ldots + \left(\frac{2}{5}\right)^n + \ldots) converges and find the sum when it exists, we can rewrite the series as a geometric series. The terms of the series seem to follow a pattern where each term is the previous term multiplied by (\left(\frac{2}{5}\right)). The series can be expressed as:

[ a_n = 6 + 19 + 3 + \frac{4}{25} + \frac{8}{125} + \frac{16}{625} + \ldots + \left(\frac{2}{5}\right)^n ]

To find the sum of this series, we first express the terms in a more general form. Notice that the first term is 6, the second term is 19 (which is (6 + 2 \cdot 6 + 2^2 \cdot 6)), the third term is 3 (which is (2^3 \cdot 6)), and so on. We can express the (n)th term as:

[ a_n = \begin{cases} 6 \cdot (2^0 + 2^1 + \ldots + 2^{k-1}), & \text{if } n = 2^k - 1 \text{ for some } k \in \mathbb{N} \ 6 \cdot 2^k, & \text{if } n = 2^k \text{ for some } k \in \mathbb{N} \end{cases} ]

Now, we can find the sum of the series by expressing it as a sum of two series: one for the terms where (n = 2^k - 1) and one for the terms where (n = 2^k).

[ \begin{aligned} S &= \sum_{k=0}^{m} \left(6 \cdot (2^0 + 2^1 + \ldots + 2^{k-1})\right) + \sum_{k=0}^{m} (6 \cdot 2^k) \ &= 6 \cdot \left(\sum_{k=0}^{m} (2^0 + 2^1 + \ldots + 2^{k-1}) + \sum_{k=0}^{m} 2^k\right) \end{aligned} ]

The first sum is a geometric series with a sum of (2^m - 1), and the second sum is a geometric series with a sum of (2^{m+1} - 1). Therefore, the sum of the series is:

[ S = 6 \cdot \left((2^m - 1) + (2^{m+1} - 1)\right) = 6 \cdot (2^m + 2^{m+1} - 2) ]

If the series converges, its sum will be the limit of the partial sums as (m) approaches infinity.