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Simplify$$\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{24} + \sqrt{25}}.$$

I know you can solve this using generating functions but I'm not totally sure.

Michael Hardy
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user68472
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  • Almost the same: http://math.stackexchange.com/questions/310962/value-of-sum-limits-k-181-dfrac1-sqrtk-sqrtk1-dfrac1 – Zander Apr 09 '13 at 09:20

2 Answers2

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Hint: Multiply top and bottom of $\dfrac{1}{\sqrt{k+1}+\sqrt{k}}$ by $\sqrt{k+1}-\sqrt{k}$, and watch the house of cards collapse.

André Nicolas
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Here, we simply use André Nicolas' hint to observe the "collapsing house of cards": We first represent the $k$th term of the sum, to simplify matters.

$$\sum_{k = 1}^{24} \color{blue}{\bf \frac{1}{\sqrt k + \sqrt{k+1}}}\quad = \quad\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{24} + \sqrt{25}}$$

$$ {\bf NOTE:}\quad \frac{1}{\sqrt k + \sqrt{k+1}}\cdot \frac{\sqrt k - \sqrt{k+1}}{\sqrt k - \sqrt{k+1}} = \frac{\sqrt k - \sqrt{k+1}}{k - (k + 1)} = \color{blue}{\bf \sqrt {k+1} - \sqrt{k}}$$ $$ $$ $$ \begin{align}\sum_{k = 1}^{24} \frac{1}{\sqrt k + \sqrt{k+1}} \quad & = \quad\sum_{k=1}^{24} \sqrt {k+1} - \sqrt{k} \\ \\ & = (\sqrt 2 - 1) + (\sqrt 3 - \sqrt 2) + (\sqrt 4 - \sqrt 3) \cdots + (\sqrt{25} -\sqrt {24}) \\ \\ & = \sqrt{25} - 1= 4 \\ \end{align} $$

amWhy
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    What does this add to the other answer? – Did Apr 07 '13 at 13:58
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    @Did I know you live in a rather lofty realm, so perhaps you forget not everyone has your level of "acuity" or "brilliance". Some people are actually just learning about these sorts of sums/series... – amWhy Apr 07 '13 at 14:06
  • I do not know how you know that. Anyway, even if this answer is a tad too rhetorical for my taste, thanks for it. If I get your point correctly, you are saying that you added this answer for the readers who failed to get the argument in André's answer. Maybe these exist. – Did Apr 07 '13 at 14:43