<s>[INST] Finding $\sum\limits_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $ How we can find
$$\sum_{n=1}^{9999}  \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})}   $$    
 [/INST] $\frac {1}{(\sqrt {n+1}+\sqrt {n})(\sqrt[4] {n+1}+\sqrt[4] {n})}=\frac {(\sqrt[4] {n+1}-\sqrt[4] {n})}{(\sqrt {n+1}+\sqrt {n})(\sqrt[4] {n+1}+\sqrt[4] {n})(\sqrt[4] {n+1}-\sqrt [4]{n})}=\frac {(\sqrt[4] {n+1}-\sqrt [4]{n})}{n+1-n}=\sqrt[4] {n+1}-\sqrt[4] {n}$ so the sum is transformed to $\sum _{n=1}^{9999} \sqrt[4] {n+1}-\sqrt[4] {n}=\sqrt[4] {10000}-\sqrt[4] {1}=9$. 
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