Is the series (8+1^n)/(1+2^n) divergent or convergent?
your numerator is 9, the denominator gets bigger, so the result approaches zero
so it is convergent.
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.
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.