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Convergence and Divergence : Limits of Sequences and Series

Each series telescopes. In each case - express the nth partial sum Sn in terms of n and determine whether the series converges or diverges.

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Solution to question 1.

a) Sn=(1/3-1/5)+(1/5-1/7)+...+[1/(2n+1)-1/(2n+3)]
=1/3-1/(2n+3)
So, lim Sn=1/3
So the series converges.

b) Since 1/[(2k-1)(2k+1)]=[1/(2k-1)-1/(2k+1)]/2, we have
Sn=(1/2-1/6)+(1/6-1/10)+...+{1/[2(2n-1)]-1/[2(2n+1)]}
=1/2-1/[2(2n+1)]
So, lim Sn=1/2
So the series converges.

c) Since [sqrt(k+1)-sqrt(k)]/sqrt(k^2+k)=1/sqrt(k)-1/sqrt(k+1), we ...

Solution Summary

Convergence and Divergence are shown in terms of the Limits of Sequences and Series.

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