Calculate two sums: $\sum_{i=1}^{99}\frac{1}{\sqrt{i+1}+\sqrt{i}}$, $\sum_{i=1}^{9999}\frac{1}{(\sqrt{i}+\sqrt{i+1}) (\sqrt[4]{i}+\sqrt[4]{i+1})}$.

Question

1. Calculate $$\sum_{i=1}^{99}\frac{1}{\sqrt{i+1}+\sqrt{i}}$$

I've figured out that the answer is 9 -there is a pattern that I've figured out. I've created a code and solved it... but how could I do it more mathematically?

2. Calculate $$\sum_{i=1}^{9999}\frac{1}{(\sqrt{i}+\sqrt{i+1}) (\sqrt[4]{i}+\sqrt[4]{i+1})}$$

This one is a total mystery, any help would be appreciated. Sorry for bad formatting. And, I almost forgot; I'm new with computing sums, please provide clear and well explained answers.

Answer 1

Hint. The first sum can be easily handled observing that $$ \frac{1}{\sqrt{i+1}+\sqrt{i}}=\sqrt{i+1}-\sqrt{i} $$ then by telescoping one gets

$$ \sum_{i=1}^n\frac{1}{\sqrt{i+1}+\sqrt{i}}=\sqrt{n+1}-1. $$

The second sum can also be easily handled observing that $$ \frac{1}{(\sqrt{i+1}+\sqrt{i})(\sqrt[4]{i+1}+\sqrt[4]{i})}=\sqrt[4]{i+1}-\sqrt[4]{i} $$ then by telescoping one gets

$$ \sum_{i=1}^n\frac{1}{(\sqrt{i+1}+\sqrt{i})(\sqrt[4]{i+1}+\sqrt[4]{i})}=\sqrt[4]{n+1}-1. $$

Answer 2

Now, we have the following: $$\sum_{i=1}^{9999} \frac{1}{(\sqrt{i}+\sqrt{i+1}) \cdot\sqrt[4]{i}+\sqrt[4]{i+1}}$$ Whenever I see these kinds of problems, I always take the conjugate of the higher degree root first since that usually leads to a simpler expression. Thus, multiply both the numerator and denominator by $\sqrt[4] i-\sqrt[4]{i+1}$: $$\sum_{i=1}^{9999} \frac{\sqrt[4] i-\sqrt[4]{i+1}}{(\sqrt{i}+\sqrt{i+1}) \cdot\sqrt {i}-\sqrt {i+1}}$$ Simplify the denominator: $$\sum_{i=1}^{9999} \frac{\sqrt[4] i-\sqrt[4]{i+1}}{i-(i+1)}$$ Simplify the denominator: $$\sum_{i=1}^{9999} \frac{\sqrt[4] i-\sqrt[4]{i+1}}{-1}$$ Divide the denominator into the numerator: $$\sum_{i=1}^{9999} \sqrt[4]{i+1}-\sqrt[4]i$$ Now, telescope: $$\sqrt[4]{9999+1}-\sqrt[4]1=\sqrt[4]{10000}-\sqrt[4]1=10-1=9$$