Want to understand how a fraction is simplified

Question

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.

Answer 1

• First equal: multiply and divide by $(k+1)\sqrt k-k\sqrt{k+1}$

• Second equal: Expand the denominator

• Third equal: cancel obvious terms in the denominator and write $k^2+k=k(k+1)$.

• Fourth equal: distribute along the minus in the numerator, and cancel the obvious factors $(k+1)$ in the first summand, and $k$ in the second one: $$ \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k(k+1)} =\frac{(k+1)\sqrt{k}}{k(k+1)}-\frac{k\sqrt{k+1}}{k(k+1)} =\frac{\sqrt k}k-\frac{\sqrt{k+1}}{k+1}. $$

• Fifth equal: $\frac{\sqrt k}{k}=\frac1{\sqrt k}$, and similarly for $k+1$.

Answer 2

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:

$$\frac{1}{ab(a+b)}=\dfrac{b-a}{ab}=\dfrac{\cancel{b}}{a\cancel{b}}-\dfrac{\bcancel{a}}{\bcancel{a}b}=\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}$$