# Want to understand how a fraction is simplified The fraction is used to determine the sum of a telescopic series. sum_(k=1)^oo 1/((k+1)sqrt(k)+k sqrt(k+1))

Want to understand how a fraction is simplified
The fraction is used to determine the sum of a telescopic series.
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{k+1}}$
This is the solved fraction.
$\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{k+1}}=\frac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{\left(k+1{\right)}^{2}k-{k}^{2}\left(k+1\right)}=\frac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{{k}^{3}+2{k}^{2}+k-{k}^{3}-{k}^{2}}=\frac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{k\left(k+1\right)}=\frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$
I want to understand what is being done on every step,mainly the last three. Thank you.
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Ashlee Ramos
First equal: multiply and divide by $\left(k+1\right)\sqrt{k}-k\sqrt{k+1}$
Second equal: Expand the denominator
Third equal: cancel obvious terms in the denominator and write ${k}^{2}+k=k\left(k+1\right)$.
Fourth equal: distribute along the minus in the numerator, and cancel the obvious factors $\left(k+1\right)$ in the first summand, and k in the second one:
$\frac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{k\left(k+1\right)}=\frac{\left(k+1\right)\sqrt{k}}{k\left(k+1\right)}-\frac{k\sqrt{k+1}}{k\left(k+1\right)}=\frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1}.$
Fifth equal: $\frac{\sqrt{k}}{k}=\frac{1}{\sqrt{k}}$, and similarly for $k+1$

tuzkutimonq4
Another way to look at it is first "declutter" the expression, since all those radicals obfuscate the simple structure. Let $a=\sqrt{k}\phantom{\rule{thinmathspace}{0ex}}$, $b=\sqrt{k+1}\phantom{\rule{thinmathspace}{0ex}}$, then:
$\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{k+1}}=\frac{1}{a{b}^{2}+{a}^{2}b}=\frac{1}{ab\left(a+b\right)}$
Now consider that $\phantom{\rule{thinmathspace}{0ex}}\left(b+a\right)\left(b-a\right)={b}^{2}-{a}^{2}=\left(\overline{)k}+1\right)-\overline{)k}=1\phantom{\rule{thinmathspace}{0ex}}$, so $\phantom{\rule{thinmathspace}{0ex}}\frac{1}{a+b}=b-a\phantom{\rule{thinmathspace}{0ex}}$. Then:
$\frac{1}{ab\left(a+b\right)}=\frac{b-a}{ab}=\frac{\overline{)b}}{a\overline{)b}}-\frac{\overline{)a}}{\overline{)a}b}=\frac{1}{a}-\frac{1}{b}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$