The tutor checks an infinite series for convergence.

Number 22, page 545, of Larson and Hostetler¹ asks about the convergence of the series

Σ₂^∞(n^-1(n²-1)^-0.5)

Solution:

First, we realize that, for n>1, n²-1 > (n-1)² Therefore,

Σ₂^∞(n^-1(n²-1)^-0.5) < Σ₂^∞(n^-1((n-1)²)^-0.5)=Σ₂^∞(n(n-1))^-1

In turn,

Σ₂^∞(n(n-1))^-1<Σ₂^∞(n-1)^-2=Σ₁^∞(n)^-2

Therefore,

Σ₂^∞(n^-1(n²-1)^-0.5) < Σ₁^∞(n)^-2

We know that Σ₁^∞(n)^-2 converges; it’s a p-series with p > 1. Therefore, being less than Σ₁^∞(n)^-2, the series in question

Σ₂^∞(n^-1(n²-1)^-0.5)

must converge also (by Limit Comparison Test).

HTH:)

¹Larson, Roland E. and Robert P. Hostetler. Calculus, third ed.
Toronto: D C Heath and Company, 1989.

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