Question

Verify that the infinite series diverges. $$ \sum_{n=1}^{\infty} \frac{2^{n}+1}{2^{n+1}} $$

Short answer

The given infinite series \( \sum_{n=1}^{\infty} \frac{2^{n}+1}{2^{n+1}} \) simplifies to \( \sum_{n=1}^{\infty} \big(\frac{1}{2} + \frac{1}{2^{n+1}}\big)\). The limit of this sequence as \(n\) approaches infinity is not zero, therefore according to the divergence test, the series diverges.

Step-by-step solution

Identify the series

The series provided is \( \sum_{n=1}^{\infty} \frac{2^{n}+1}{2^{n+1}} \). This is an infinite series because it extends to infinity.

Simplify the series

The series can be rewritten by combining like terms in the numerator. Divide both terms in the numerator by \(2^{n+1}\), giving the simplified series \( \sum_{n=1}^{\infty} \big(\frac{2^{n}}{2^{n+1}} + \frac{1}{2^{n+1}}\big) \). This simplifies further to \( \sum_{n=1}^{\infty} \big(\frac{1}{2} + \frac{1}{2^{n+1}}\big)\). This series clearly separates into two series: one is a constant and the other is a geometric series with a ratio of \( 1/2 \).

Evaluate the limit of the sequence

Consider the limit as \(n\) approaches infinity for the sequence \( a_n = \frac{1}{2} + \frac{1}{2^{n+1}} \). The second term vanishes as \( n \) approaches infinity, hence the limit will be \( \frac{1}{2} \), which is not equal to zero.

Apply the divergence test

Since the limit of the sequence of terms of the series is not zero, by the divergence test, the infinite series is divergent.