Is the series (8+1^n)/(1+2^n) divergent or convergent?

User Icon for Reiny Reiny answered
15 years ago

your numerator is 9, the denominator gets bigger, so the result approaches zero

so it is convergent.

User Icon for Shaniquaa Shaniquaa answered
15 years ago

that doesn't mean it's convergent, necessarily.

but i found it is actually convergent because if you divide by 1/2^n, which is also a positive term series, you get 9, which means the series is convergent.

User Icon for Explain Bot Explain Bot answered
11 months ago

To determine whether the series (8 + 1^n) / (1 + 2^n) is divergent or convergent, we need to analyze its behavior as n approaches infinity.

The given series can be rewritten as (8 / (1 + 2^n)) + (1^n / (1 + 2^n)). Let's examine each term separately.

1. The term 8 / (1 + 2^n):
As n approaches infinity, the term 2^n grows much faster than 1. Therefore, as n approaches infinity, the denominator of this term becomes significantly larger, making the whole term approach zero. In other words, 8 / (1 + 2^n) approaches 8 / ∞, which is equal to 0.

2. The term 1^n / (1 + 2^n):
Since any number raised to the power of n where n is approaching infinity will either tend to 1 (if the number is equal to 1) or 0 (if the number is less than 1), the term 1^n / (1 + 2^n) approaches 1 / (1 + ∞), which simplifies to 1 / ∞ and equals to 0 as well.

As both terms of the series approach zero, the sum of the series is the sum of two zeros, which is equal to zero.

Therefore, the series (8 + 1^n) / (1 + 2^n) is convergent, with the sum equal to 0.