Question

\[n\in \mathbb{N}\]
Show that :
\[\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}\]
Tipp: expand with \[(\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}})\]

Answer 1

hard to prove something that isn't true

Answer 2

i am sorry, i wrote wrong

Answer 3

take the left hand side and make it into one fraction. Next, multiply it by your 'tip' and simplify

Answer 4

i found left side \[\frac{1}{2n+1+2n\sqrt{1}}\] but i dont know somewhat it feels wrong

Answer 5

\[\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\]

\[=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}\]

Answer 6

\[=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}\times\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\]

\[=\frac{(n+1)-(n)}{\sqrt{n}\sqrt{n+1}\sqrt{n+1}+\sqrt{n}\sqrt{n+1}\sqrt{n}}\]

\[=\frac{1}{\sqrt{n}(n+1)+n\sqrt{n+1}}\]

Answer 7

which is equivalent to what you have above (using commutativity)

Answer 8

thx Zarkon, now i get it