Answer :
Step-by-step explanation:
To determine whether the given alternating series converges:
\[ \sum_{k=1}^{\infty} \frac{(-1)^{k+1} (k+3)}{2k+3} \]
we can use the Alternating Series Test (Leibniz's Test), which states that an alternating series \(\sum (-1)^k a_k\) converges if:
1. The absolute value of the terms \(a_k\) is decreasing.
2. The limit of \(a_k\) as \(k\) approaches infinity is 0.
Let's check these conditions for the given series.
### 1. Check if the terms are decreasing:
We need to examine whether \(\frac{k+3}{2k+3}\) is decreasing.
Compare \(a_k\) and \(a_{k+1}\):
\[ a_k = \frac{k+3}{2k+3} \]
\[ a_{k+1} = \frac{(k+1)+3}{2(k+1)+3} = \frac{k+4}{2k+5} \]
We need to check if \(a_k > a_{k+1}\):
\[ \frac{k+3}{2k+3} > \frac{k+4}{2k+5} \]
Cross-multiplying to compare the fractions:
\[ (k+3)(2k+5) > (k+4)(2k+3) \]
\[ 2k^2 + 5k + 6k + 15 > 2k^2 + 3k + 8k + 12 \]
\[ 2k^2 + 11k + 15 > 2k^2 + 11k + 12 \]
\[ 15 > 12 \]
This inequality holds true, indicating that \(a_k\) is decreasing for all \(k\).
### 2. Check if the limit of \(a_k\) is 0:
We need to find \(\lim_{k \to \infty} \frac{k+3}{2k+3}\):
\[ \lim_{k \to \infty} \frac{k+3}{2k+3} = \lim_{k \to \infty} \frac{1 + \frac{3}{k}}{2 + \frac{3}{k}} \]
As \(k\) approaches infinity, the terms \(\frac{3}{k}\) go to 0:
\[ \lim_{k \to \infty} \frac{1 + \frac{3}{k}}{2 + \frac{3}{k}} = \frac{1 + 0}{2 + 0} = \frac{1}{2} \]
Since the limit is \(\frac{1}{2}\) and not 0, the second condition of the Alternating Series Test is not met.
