How do you find the sum $$\sum_{k = 1}^n \frac{1}{(k+1)\sqrt{k}+k\sqrt{(k+1)}}?$$ Sorry for not provide any of my idea since I have no idea about what to do.
I hope everyone here could help me. Thank you
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Are there parentheses somewhere? Is the first $\sqrt k$ in the numerator or denominator? To get square root signs in MathJax use \sqrt and put the argument in braces to get it under the square root sign. You could get a good approximation from converting it to an integral. – Ross Millikan Nov 30 '19 at 14:58
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1I edited the expression to make it more readable, please tell me if this is correct. – stochasticboy321 Nov 30 '19 at 15:00
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1If it is correct - try rationalising the denominator term-by-term by multiplying and dividing by $(k+1)\sqrt{k} - k\sqrt{(k+1)}$. You end up with a telescoping sum. – stochasticboy321 Nov 30 '19 at 15:01
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@stochasticboy321 yes it is correct. Sorry for my trouble – user635988 Nov 30 '19 at 15:09
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As suggested in the comments, we have that
$$ \frac{1}{(k+1)\sqrt{k}+k\sqrt{(k+1)}}\cdot \frac{(k+1)\sqrt{k}-k\sqrt{(k+1)}}{(k+1)\sqrt{k}-k\sqrt{(k+1)}}= \frac{(k+1)\sqrt{k}-k\sqrt{(k+1)}}{k^2+k}=$$
$$=\frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$$
and we obtain a telescoping sum.
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