1.For n∈Nn \in \mathbb{N}n∈N, let an=∑k=1n2ka_n = \sum_{k=1}^{n} 2kan=∑k=1n2k and bn=∑k=1n(2k−1)b_n = \sum_{k=1}^{n} (2k-1)bn=∑k=1n(2k−1). Then limn→∞(an−bn)\displaystyle \lim_{n \to \infty} \left(\sqrt{a_n} - \sqrt{b_n}\right)n→∞lim(an−bn) is equal toa.111b.12\displaystyle \frac{1}{2}21c.000d.222Login to continueOnly logged in users canattempt or see the solution.
