How can you change this series into a telescoping series so then you can find its sum? $$\sum_{n=1}^{\infty} \frac{1}{\sqrt n + \sqrt{n+1}}$$
Sum of telescoping series
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$\begingroup$
convergence
summation
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0Just a general check before you proceed: Is this series convergent? – 2017-01-26
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0I do not think so. – 2017-01-26
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0no it does not converge do to a limit comparison test with $1/\sqrt n$ – 2017-01-26
1 Answers
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Multiply by conjugate on top and bottom.
$$\frac{1}{\sqrt n + \sqrt{n+1}} \frac{\sqrt n -\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}}$$
$$=\sqrt{n+1}-\sqrt{n}$$
So,
$$\sum_{n=1}^{N} \left(\sqrt{n+1}-\sqrt{n} \right)=\sqrt{N+1}-\sqrt{1}$$
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0so the denominator all cancels out? – 2017-01-26
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2which shows that the series does not converge, but has nice partial sum, $\sum\limits_{i=1}^n\frac{1}{\sqrt{i}+\sqrt{i+1}}=\sqrt{n+1}-1$ – 2017-01-26
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0In the denominator we have the difference of two squares. $(\sqrt{n})^2-(\sqrt{n+1})^2=n-(n+1)=-1$ @bjp409 – 2017-01-26