Consider the series 1/4+1/6+1/9+2/27+4/81+....

Does the series converge or diverge? Is the series arithmetic, geometric, neither, or geometric with an absolute value of the common ration being greater/ less than 1?

Does the series converge or diverge? If it converges, what is the sum? Show your work.

∑∞n=1−4(−1/3)n−1

This is a geometric series with first term a=-4 and common ratio r=-1/3. In order for it to converge, we need to have |r| < 1, which is the case here.

The sum of an infinite geometric series with first term a and common ratio r, where |r| < 1, is given by:

S = a / (1-r)

Plugging in the values a=-4 and r=-1/3, we get:

S = (-4) / [1 - (-1/3)]
S = (-4) / (4/3)
S = -3

Therefore, the series converges and its sum is -3.

To determine if the series converges or diverges and to identify the type of series, we need to analyze its pattern.

The given series is: 1/4 + 1/6 + 1/9 + 2/27 + 4/81 + ...

We can see that each term of the series is the fraction 1/(3^n) multiplied by 2 raised to the power of (n-2).

To simplify the series, let's express it in terms of the exponent n:

1/4 + 1/6 + 1/9 + 2/27 + 4/81 + ... = (1/3^2)*(2^0) + (1/3^3)*(2^1) + (1/3^4)*(2^2) + (1/3^5)*(2^3) + (1/3^6)*(2^4) + ...

Now, notice that the terms can be written as (1/3^n)*(2^(n-2)).

We can split each term into two separate parts: (1/3^n) and (2^(n-2)), then rewrite the series as the product of two separate series:

(1/3^2)*(2^0) + (1/3^3)*(2^1) + (1/3^4)*(2^2) + (1/3^5)*(2^3) + (1/3^6)*(2^4) + ...
= [(1/3^2) + (1/3^3) + (1/3^4) + (1/3^5) + (1/3^6) + ...] * [(2^0) + (2^1) + (2^2) + (2^3) + (2^4) + ...]

The first part of the product represents a geometric series with a common ratio of 1/3, and the second part is a geometric series with a common ratio of 2.

Now, let's determine if each series converges or diverges:

1. The first geometric series with a common ratio of 1/3 will converge since the absolute value of the common ratio (|1/3|) is less than 1.

2. The second geometric series with a common ratio of 2 will diverge since the absolute value of the common ratio (|2|) is greater than 1.

Since the product of a convergent series and a divergent series can only result in a divergent series, we can conclude that the given series diverges.

In summary, the series diverges and is neither arithmetic nor geometric.

since r = 2/3 it will converge