2006 SMT/Team Problems/Problem 8

Solution

To begin, let's rewrite the sum as

$\lim_{n\to\infty} \sum_{k=n^2}^{(n+1)^2} \frac{1}{\sqrt{k}} = \lim_{n\to\infty} \sum_{k=0}^{2n+1} \frac{1}{\sqrt{n^2+k}}$

By the Squeeze Theorem:

$\begin{align*} \lim_{n\to\infty} \sum_{k=0}^{2n+1} \frac{1}{\sqrt{n^2}} &\ge\lim_{n\to\infty} \sum_{k=0}^{2n+1} \frac{1}{\sqrt{n^2+k}} \ge \lim_{n\to\infty} \sum_{k=0}^{2n+1} \frac{1}{\sqrt{n^2+2n+1}}\\ \lim_{n\to\infty} \sum_{k=0}^{2n+1} \frac{1}{n} &\ge\lim_{n\to\infty} \sum_{k=0}^{2n+1} \frac{1}{\sqrt{n^2+k}} \ge \lim_{n\to\infty}\sum_{k=0}^{2n+1} \frac{1}{n+1} \\ \lim_{n\to\infty} \frac{2n+2}{n} &\ge\lim_{n\to\infty} \sum_{k=0}^{2n+1} \frac{1}{\sqrt{n^2+k}} \ge \lim_{n\to\infty} \frac{2n+2}{n+1} \\ \lim_{n\to\infty} 2+\frac{2}{n} &\ge\lim_{n\to\infty} \sum_{k=0}^{2n+1} \frac{1}{\sqrt{n^2+k}} \ge 2 \\ 2&\ge\lim_{n\to\infty} \sum_{k=0}^{2n+1} \frac{1}{\sqrt{n^2+k}} \ge 2 \end{align*}$

Therefore, $\lim_{n\to\infty} \sum_{k=0}^{2n+1}\frac{1}{\sqrt{n^2+k}}=\lim_{n\to\infty} \sum_{k=n^2}^{(n+1)^2} \frac{1}{\sqrt{k}} = \boxed{2}$

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2006 SMT/Team Problems/Problem 8

Solution