Does 3/(2^(n) +1) diverge or converge in an infinite sum?
Do the zero limit test.
even without that, you know that 1/2^n converges. It's just a geometric series.
1/(2^n+1) is even less. The 3 is just noise.
To determine whether the series sum of 3/(2^n + 1) converges or diverges, we can use the concept of convergence and divergence of series.
First, let's rewrite the series in a familiar form:
3/(2^n + 1) = 3 * (1/(2^n + 1))
Now, we can analyze the behavior of the series by examining its terms as n approaches infinity.
As n tends toward infinity, 1/(2^n + 1) approaches 0 since the denominator becomes extremely large. This means that the terms of the series get closer and closer to 0 as n increases.
Next, we need to check if the series converges or diverges. For this, we can use the comparison test. Let's select a known convergent geometric series as our comparison series.
Consider the geometric series: 1/2^n
Here, the common ratio is 1/2, and we know that this series converges.
Now, let's compare the terms of our original series (3/(2^n + 1)) with the terms of the convergent geometric series (1/2^n):
3/(2^n + 1) < 3/(2^n) = 1.5 * (1/2^n)
Since 1.5 * (1/2^n) is a convergent geometric series, and our original series 3/(2^n + 1) is less than it, we can conclude that 3/(2^n + 1) also converges.
Therefore, the series sum of 3/(2^n + 1) converges.