Want to understand how a fraction is simplified The fraction is used to determine the sum of a telescopic series.
$$\sum_{k=1}^\infty\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}$$
This is the solved fraction.
$$\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1)^2k-k^2(k+1)} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k^3+2k^2+k-k^3-k^2} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k(k+1)} = \frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1} = \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$$
I want to understand what is being done on every step,mainly the last three.
Thank you.

Another way to look at it is first "declutter" the expression, since all those radicals obfuscate the simple structure. Let $a=\sqrt{k}\,$, $b=\sqrt{k+1}\,$, then:
$$\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}} = \frac{1}{ab^2+a^2b}=\frac{1}{ab(a+b)}$$
Now consider that $\require{cancel}\,(b+a)(b-a)=b^2-a^2=(\cancel{k}+1)-\cancel{k}=1\,$, so $\,\dfrac{1}{a+b}=b-a\,$. Then:
$$\frac{1}{ab(a+b)}=\dfrac{b-a}{ab}=\dfrac{\cancel{b}}{a\cancel{b}}-\dfrac{\bcancel{a}}{\bcancel{a}b}=\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}$$
