#### Answer

$\dfrac {25}{12}$

#### Work Step by Step

$\sum ^{4}_{j=1}j^{-1}=\dfrac {1}{1}+\dfrac {1}{2}+\dfrac {1}{3}+\dfrac {1}{4}=\dfrac {25}{12}$

Published by
Pearson

ISBN 10:
013421742X

ISBN 13:
978-0-13421-742-0

$\dfrac {25}{12}$

$\sum ^{4}_{j=1}j^{-1}=\dfrac {1}{1}+\dfrac {1}{2}+\dfrac {1}{3}+\dfrac {1}{4}=\dfrac {25}{12}$

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