Question

Which of the following is a convergent series?

• A. \( \sum_{n=0}^{\infty} \frac{3n^2+5n+6}{5n^2+6n+3} \)
• B. \( \sum_{n=0}^{\infty} (-1)^n \frac{2n+3}{2n} \)
• C. \( \sum_{n=0}^{\infty} \frac{2^n-2}{2^n+1} \)
• D. \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\log(n+1)} \)

Hint:

• Term Test for Divergence: If \(\lim_{n \to \infty} a_n \neq 0\), then \(\sum a_n\) diverges.
• Alternating Series Test: If \(a_n = (-1)^n b_n\) or \(a_n = (-1)^{n+1} b_n\) with \(b_n>0\), then \(\sum a_n\) converges if (1) \(\lim_{n \to \infty} b_n = 0\) AND (2) \(b_n\) is a decreasing sequence.

Answer 1

The Correct Option is D

A necessary condition for a series \(\sum a_n\) to converge is that \(\lim_{n \to \infty} a_n = 0\).
If this limit is not zero, the series diverges (Term Test for Divergence). (a) \(a_n = \frac{3n^2+5n+6}{5n^2+6n+3}\). \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{3+5/n+6/n^2}{5+6/n+3/n^2} = \frac{3}{5} \neq 0\). 
Diverges by Term Test. (b) \(a_n = (-1)^n \frac{2n+3}{2n}\). \(\lim_{n \to \infty} \frac{2n+3}{2n} = \lim_{n \to \infty} \frac{2+3/n}{2} = \frac{2}{2} = 1\). So, \(\lim_{n \to \infty} a_n\) does not exist (oscillates between approx 1 and -1). Not 0. 
Diverges by Term Test. (c) \(a_n = \frac{2^n-2}{2^n+1}\). \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1-2/2^n}{1+1/2^n} = \frac{1-0}{1+0} = 1 \neq 0\). 
Diverges by Term Test. (d) \(a_n = \frac{(-1)^{n+1}}{\log(n+1)}\). 
This is an alternating series. Let \(b_n = \frac{1}{\log(n+1)}\). 
1. \(\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{1}{\log(n+1)} = 0\). (As \(n \to \infty, \log(n+1) \to \infty\)). 2. Is \(b_n\) decreasing? \(\log(x)\) is an increasing function for \(x>0\). 
So \(\log(n+1)\) is increasing for \(n \ge 1\). 
Therefore, \(b_n = \frac{1}{\log(n+1)}\) is a decreasing sequence for \(n \ge 1\). (\(\log(n+2) > \log(n+1) \implies \frac{1}{\log(n+2)} < \frac{1}{\log(n+1)} \implies b_{n+1} < b_n\)). 
Since both conditions of the Alternating Series Test are met (\(\lim_{n \to \infty} b_n = 0\) and \(b_n\) is decreasing), the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\log(n+1)} \) converges. (It converges conditionally, not absolutely, as \(\sum 1/\log(n+1)\) diverges by comparison with \(\sum 1/n\)). \[ \boxed{\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\log(n+1)}} \]