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How do you test the series #Sigma 1/sqrt(n(n+1))# from n is #[1,oo)# for convergence?

Answer 1

To test the convergence of the series ( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n(n+1)}} ), we can use the limit comparison test. Let's consider the series ( \sum_{n=1}^{\infty} \frac{1}{n} ) as a comparison.

Taking the limit as ( n ) approaches infinity of ( \frac{a_n}{b_n} ), where ( a_n = \frac{1}{\sqrt{n(n+1)}} ) and ( b_n = \frac{1}{n} ), we get:

[ \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n(n+1)}}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{\sqrt{n(n+1)}} = \lim_{n \to \infty} \sqrt{\frac{n}{n+1}} = 1 ]

Since the limit is a finite nonzero constant, and ( \sum_{n=1}^{\infty} \frac{1}{n} ) is a known divergent series (harmonic series), then by the limit comparison test, ( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n(n+1)}} ) also diverges.

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Answer 2

Cameron Alexander

See below.

#1/(n+1) < 1/sqrt(n(n+1)) < 1/n#

and

#sum 1/n# does not converge

#sum 1/sqrt(n(n+1)) # does not converge!

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.