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How do you test the series #sum_(n=1)^oo n/(n^2+2)# for convergence?

Answer 1

To test the series ∑(n=1 to ∞) n/(n^2 + 2) for convergence, we can use the Limit Comparison Test. Let's compare it with the series ∑(n=1 to ∞) 1/n.

Taking the limit as n approaches infinity of the ratio of the nth terms of the two series gives: lim (n→∞) [(n/(n^2 + 2)) / (1/n)] = lim (n→∞) [(n^2) / (n^2 + 2)] = 1.

Since the limit is a finite nonzero value, and the harmonic series ∑(n=1 to ∞) 1/n diverges, by the Limit Comparison Test, the given series also diverges.

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Answer 2

Evelyn Agard

This series diverges by the integral test.

Note that:

#x = 1/2 d/(dx) (x^2+2)#

So:

#int x/(x^2+2) dx = 1/2 ln abs(x^2+2) + C -> oo# as #x->oo#

So #sum_(n=1)^oo n/(n^2+2)# diverges by the integral test.

Alternatively, note that:

#sum_(n=1)^oo n/(n^2+2) = sum_(n=1)^oo 1/(n+2/n) >= sum_(n=1)^oo 1/(n+2) = sum_(n=3)^oo 1/n#

which diverges.

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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.