Question

If \[ u_n=\frac{\sqrt{2n^2-5n+1}}{4n^3-7n^2+2} \] and \[ v_n=\sqrt{n^2+1}-\sqrt{n^2-1}, \] then

• A. \(\displaystyle\sum_{n=1}^{\infty}u_n\) convergent and \(\displaystyle\sum_{n=1}^{\infty}v_n\) divergent
• B. \(\displaystyle\sum_{n=1}^{\infty}u_n\) and \(\displaystyle\sum_{n=1}^{\infty}v_n\), both converge
• C. \(\displaystyle\sum_{n=1}^{\infty}u_n\) divergent and \(\displaystyle\sum_{n=1}^{\infty}v_n\) convergent
• D. \(\displaystyle\sum_{n=1}^{\infty}u_n\) and \(\displaystyle\sum_{n=1}^{\infty}v_n\), both diverge

Hint:

Use asymptotic comparison: \[ \boxed{ u_n\sim\frac1{n^2}\Rightarrow\text{Convergent}, \qquad v_n\sim\frac1n\Rightarrow\text{Divergent}. } \]

Answer 1

The Correct Option is A

Step 1: Test the series involving \(u_n\). For large \(n\), \[ \sqrt{2n^2-5n+1}\sim \sqrt2\,n, \] and \[ 4n^3-7n^2+2\sim4n^3. \] Hence, \[ u_n\sim\frac{\sqrt2\,n}{4n^3} =\frac{\sqrt2}{4n^2}. \] Since \[ \sum\frac1{n^2} \] is a convergent \(p\)-series \((p=2)\), \[ \boxed{\sum u_n\ \text{converges}.} \]

Step 2: Test the series involving \(v_n\). Rationalizing, \[ v_n = \frac{(n^2+1)-(n^2-1)} {\sqrt{n^2+1}+\sqrt{n^2-1}} = \frac{2} {\sqrt{n^2+1}+\sqrt{n^2-1}}. \] For large \(n\), \[ v_n\sim\frac{2}{2n} =\frac1n. \] Since the harmonic series \[ \sum\frac1n \] diverges, \[ \boxed{\sum v_n\ \text{diverges}.} \] Therefore, \[ \boxed{(A)} \] is the correct answer.