Question

A convergent series from the given series is

• A. \[ \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}} \]
• B. \[ \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n+1}} \]
• C. \[ \sum_{n=1}^{\infty}\frac{3n^4+5}{n^2(n^2+4n+5)} \]
• D. \[ \sum_{n=3}^{\infty}\frac{\log n}{n} \]

Hint:

Remember: \[ \boxed{ \begin{aligned} \sum\frac1{n^p} &\text{ converges if }p>1, \sum(-1)^na_n &\text{ converges if }a_n\downarrow0. \end{aligned} } \]

Answer 1

The Correct Option is B

Step 1: Test each series for convergence. For option (A), \[ \sum\frac{1}{\sqrt n} = \sum\frac{1}{n^{1/2}} \] is a \(p\)-series with \[ p=\frac12<1, \] so it diverges. For option (B), \[ \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n+1}} \] is an alternating series. Since \[ \frac1{\sqrt{n+1}} \] is positive, decreasing and \[ \lim_{n\to\infty}\frac1{\sqrt{n+1}}=0, \] the series converges by the Leibniz Alternating Series Test. For option (C), \[ \frac{3n^4+5}{n^2(n^2+4n+5)} \sim 3, \] which does not approach zero. Hence the series diverges. For option (D), \[ \sum\frac{\log n}{n} \] diverges by the Integral Test.

Step 2: Choose the convergent series. Only option (B) satisfies the convergence criterion. Therefore, \[ \boxed{ \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n+1}} } \] is the convergent series. Thus, \[ \boxed{(B)} \] is the correct answer.