Evaluate the sum S for the series: S = sum from n=2

Evaluate the sum $S$ for the series: $$ S = \sum_{n=2}^{\infty} \frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}, $$ where the positive numbers $a_n$ form an arithmetic progression.

Step 1 · Properties of the arithmetic progression

Let the progression have first term $a_1$ and common difference $d$, so $a_n - a_{n-1} = d$. Since all terms $a_n$ are positive, $a_1 > 0$.

Step 2 · Decompose the general term

The general term, for $n\ge2$, is $b_n = \dfrac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$. Rationalizing by the conjugate $\sqrt{a_n}-\sqrt{a_{n-1}}$:

$$ b_n = \frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\cdot\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{\sqrt{a_n}-\sqrt{a_{n-1}}} = \frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{a_n-a_{n-1}}. $$

Since $a_n - a_{n-1}=d$ (assuming $d\neq0$):

$$ b_n = \frac{1}{d}\left(\sqrt{a_n}-\sqrt{a_{n-1}}\right), $$

a telescoping difference of consecutive terms.

Step 3 · Determine the partial sum $S_N$

Summing from $n=2$ to $N$:

$$ S_N = \frac1d \sum_{n=2}^{N} \left(\sqrt{a_n}-\sqrt{a_{n-1}}\right). $$

Writing out the terms, all interior pieces cancel:

$$ S_N = \frac1d\Big[\left(\sqrt{a_2}-\sqrt{a_1}\right)+\left(\sqrt{a_3}-\sqrt{a_2}\right)+\cdots+\left(\sqrt{a_N}-\sqrt{a_{N-1}}\right)\Big] = \frac{1}{d}\left(\sqrt{a_N}-\sqrt{a_1}\right). $$

Step 4 · Calculate the sum $S$

The sum $S$ is the limit of $S_N$ as $N \to \infty$, with $a_N = a_1+(N-1)d$:

$$ S = \lim_{N\to\infty} S_N = \lim_{N\to\infty} \frac1d\left(\sqrt{a_N}-\sqrt{a_1}\right). $$

• Case $d>0$ (increasing): $a_N \to \infty$, so $S = \frac1d(\infty-\sqrt{a_1}) = \infty$; the series diverges.
• Case $d<0$ (decreasing): since all $a_n>0$ must hold for every $n$, an infinite decreasing progression would eventually force $a_N<0$ — a contradiction. So no infinite positive arithmetic progression has $d<0$.
• Case $d=0$ (constant): $a_n=a_1$ for all $n$, so $b_n = \frac{1}{2\sqrt{a_1}}$ is a nonzero constant; summing infinitely many equal positive terms gives $S=\infty$.

Since every admissible infinite positive arithmetic progression has $d \ge 0$, the series diverges in every valid case.

Summary Table

Additional Notes

• This problem highlights a subtlety: the positivity constraint on $a_n$ rules out $d<0$ for an infinite progression, so only $d\ge0$ need be considered — and both remaining cases diverge.
• The rationalization trick, $\frac{1}{\sqrt{x}+\sqrt{y}} = \frac{\sqrt{x}-\sqrt{y}}{x-y}$, is the same idea used whenever a sum or difference of square roots appears in a denominator; it converts an awkward expression into a telescoping difference.
• The closed form $S_N = \frac1d(\sqrt{a_N}-\sqrt{a_1})$ is valid and finite for every finite $N$; it is only the limit as $N\to\infty$ that fails to exist as a finite number.