91Ó°ÊÓ

Let the series be denoted by \(S\). To find the general term, we need to identify the pattern in the series. We can describe the terms in the series by the following function: \[ S_n = \frac{(1 + \sqrt{n+1})}{2(n)} \] where \(n\) starts from 1. This series is alternating since the terms have alternating signs.

We need to show that \(b_n \geq b_{n+1}\), where \(b_n = \frac{(1 + \sqrt{n+1})}{2(n)}\). To prove this, let's compare \(b_n\) and \(b_{n+1}\): \[ \frac{(1 + \sqrt{n+1})}{2(n)} \stackrel{?}{\geq} \frac{(1 + \sqrt{n+2})}{2(n+1)} \] By cross-multiplying, we get: \[ (1 + \sqrt{n+1})(2n + 2) \stackrel{?}{\geq} (1 + \sqrt{n+2})(2n) \] Simplify and solve: \[ 2n + 2 + 2n\sqrt{n+1} + 2\sqrt{n+1} \stackrel{?}{\geq} 2n + 2\sqrt{n+2} + 2n\sqrt{n+1} \] We can cancel out the common terms on both sides: \[ 2\sqrt{n+1} \stackrel{?}{\geq} 2\sqrt{n+2} \] Divide both sides by 2: \[ \sqrt{n+1} \stackrel{?}{\geq} \sqrt{n+2} \] Since the square root function is monotonically increasing, this inequality is not true. Thus, the terms of the series do not decrease monotonically.

We need to find the limit of the terms as \(n\) goes to infinity: \[ \lim_{n \to \infty} \frac{(1 + \sqrt{n+1})}{2(n)} \] To find this limit, we can use L'Hopital's Rule since this is an indeterminate form \(\frac{\infty}{\infty}\). First, differentiate the numerator and denominator with respect to \(n\): \[ \lim_{n \to \infty} \frac{\frac{1}{2\sqrt{n+1}}}{2} \] Simplify the expression: \[ \lim_{n \to \infty} \frac{1}{4\sqrt{n+1}} \] As \(n\) goes to infinity, the limit of the terms approaches zero: \[ \lim_{n \to \infty} \frac{1}{4\sqrt{n+1}} = 0 \]

Although the limit of the terms of the series approaches zero, the terms of the series do not decrease monotonically as we proved in Step 2. Since both conditions of the Alternating Series Test are not met, we cannot conclude that this series converges.