# Simplifying a finite summation expression.

I found a question that asks to evaluate the following expression:

$$\frac{1}{\sqrt4 +\sqrt5} + \frac{1}{\sqrt5 +\sqrt6} + \frac{1}{\sqrt6 +\sqrt7} + … + \frac{1}{\sqrt{624} +\sqrt{625}}$$

I was able to represent the above expression in summation notation as such:

$$\sum_{k=4}^{624}\frac{1}{\sqrt k +\sqrt{k+1}}$$

Rationalizing the denominator of $$\frac{1}{\sqrt k +\sqrt{k+1}}$$ gave me $$\sqrt{k+1} - \sqrt k$$.

So then the new summation expression would be:

$$\sum_{k=4}^{624}\sqrt{k+1} - \sqrt k = \sum_{k=4}^{624}\sqrt{k+1} - \sum_{k=4}^{624}\sqrt k$$

However, after this point, I'm stuck. I'm unsure of how exactly to proceed and simplify the expression. Any help is appreciated.

• Telescoping series. Usually, 2 terms remain after cancelations. Usually, the greatest and the smallest. I guess the answer is $\sqrt{625}-\sqrt{4}=25-2=23$. Oct 16, 2022 at 7:24
• Mathjax: Use braces {} instead of parentheses around $k+1$ and you properly get $\sqrt{k+1}$. Oct 16, 2022 at 8:45

\begin{align} \sum_{n=4}^{624}\frac{1}{\sqrt{k}+\sqrt{k+1}}= &=\sum_{n=4}^{624}\frac{1}{\sqrt{k+1}+\sqrt{k}}\\ &=\sum_{n=4}^{624}\frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}\\ &=\sum_{n=4}^{624}(\sqrt{k+1}-\sqrt{k})\\ &=\sum_{n=4}^{624}\sqrt{k+1}-\sum_{n=4}^{624}\sqrt{k}\\ &=\sum_{n=5}^{625}\sqrt{k}-\sum_{n=4}^{624}\sqrt{k}\\ &=\sqrt{625}+\sum_{n=5}^{624}\sqrt{k}-\sum_{n=5}^{624}\sqrt{k}-\sqrt{4}\\ &=25-2\\ &=23\\ \end{align}
Your sum can be evaluated as a so-called telescoping series. For instance, $$\sum_{n=1}^{N-1} (a_{n+1}-a_{n}) = a_N - a_1$$. Convince yourself of this, or you can prove it via induction.