Porisma

Worked solution

Rationalise a radical sum and distinguish terms from partial sums

Given ∑n=1N(n+1−n−1)=N+N+1−1\sum_{n=1}^N(\sqrt{n+1}-\sqrt{n-1})=\sqrt N+\sqrt{N+1}-1:

(i) Find exactly (a) ∑n=50100(n+1−n−1)\sum_{n=50}^{100}(\sqrt{n+1}-\sqrt{n-1}) marks, and (b) ∑n=501001/(n+1+n−1)\sum_{n=50}^{100}1/(\sqrt{n+1}+\sqrt{n-1}). mark

(ii) State with a reason whether ∑n=1∞(n+1−n−1)\sum_{n=1}^{\infty}(\sqrt{n+1}-\sqrt{n-1}) converges. mark

[2013/PJC/Promo/8(modified)]

CJC · Review problems · Question 6

Answer

(a) 3−52+1013-5\sqrt2+\sqrt{101}; (b) half that value; the infinite series diverges.

(i)(a) Subtract the partial sum through 49

The sum is (100+101−1)−(49+50−1)=3−52+101(\sqrt{100}+\sqrt{101}-1)-(\sqrt{49}+\sqrt{50}-1)=3-5\sqrt2+\sqrt{101}.

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(i)(b) Multiply by the conjugate

1/(n+1+n−1)=(n+1−n−1)/[(n+1)−(n−1)]=(n+1−n−1)/21/(\sqrt{n+1}+\sqrt{n-1})=(\sqrt{n+1}-\sqrt{n-1})/[(n+1)-(n-1)]=(\sqrt{n+1}-\sqrt{n-1})/2. Thus the sum is 12(3−52+101)\tfrac12(3-5\sqrt2+\sqrt{101}).

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(ii) Test the partial sums, not only the individual terms

As N→∞N\to\infty, N+N+1−1→∞\sqrt N+\sqrt{N+1}-1\to\infty. Therefore the series diverges. Its terms tend to zero, but that condition alone does not imply convergence.

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Common mistakes

A term tending to zero does not prove its infinite series converges.

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