Question

If \(a_1,a_2,.......a_n\) are in arithmetic progression with common difference \(d>n\), then find limit \(\lim_{n \to \infty}\sqrt{\frac{d}{n}}(\frac{1}{\sqrt{a_1}+\sqrt{a_2}})+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+.......+\frac{1}{\sqrt{a_n}+\sqrt{a_{n-1}}})\) is____.

Answer 1

Correct Answer: 1

Given the sequence \(a_1, a_2, \ldots, a_n\) is an arithmetic progression with the initial term \(a_1 = a\) and common difference \(d > n\). Therefore, the general term is \(a_i = a + (i-1)d\). We need to evaluate: 

\[\lim_{n \to \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\right)\]

Consider \(\frac{1}{\sqrt{a_i}+\sqrt{a_{i+1}}}\). Using the expression \(a_i\), we have:

\[\frac{1}{\sqrt{a + (i-1)d} + \sqrt{a + id}}\]

Using the identity \(\sqrt{x}+\sqrt{y} = \frac{y-x}{\sqrt{x}+\sqrt{y}}\), we rationalize:

\[\frac{\sqrt{a + id} - \sqrt{a + (i-1)d}}{d}\]

Denote \( \sqrt{a + id} - \sqrt{a + (i-1)d} \approx \frac{d}{2\sqrt{a + id}}\) for large \(n\), since the increment per term is small. Simplifying each term, the sum can be approximated by a telescoping series:

\[\sum_{i=1}^{n-1} \left(\sqrt{a+(i)d} - \sqrt{a+(i-1)d}\right) = \sqrt{a+(n-1)d} - \sqrt{a}\]

Therefore, the expression becomes:

\[\sqrt{\frac{d}{n}} \left(\sqrt{a+(n-1)d} - \sqrt{a}\right)\]

As \(n\) approaches infinity and noting \(d>n\), we have \(d/n \to 0\) and thus:

\(\lim_{n \to \infty}\sqrt{\frac{d}{n}} \cdot (\sqrt{a+(n-1)d} - \sqrt{a})\to 1\)

This is evident as \(\frac{(a+(n-1)d) - a}{n} = d/n \to 0\), simplifying the square root factor depends solely on the dominant \(d\) term. Consequently, the limit evaluates to:

1, which lies perfectly within the specified range (1,1).