I am not sure

Calculus Level 2

S = 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 + ⋯ S=1+\dfrac1{2}+\dfrac1{3}+\dfrac1{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+ \cdots

Find the value of S S as defined above.

Select 666 if you come to the conclusion that the series fails to converge

15 12 10 666

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Sabhrant Sachan by Sabhrant Sachan , 21, India

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2 solutions

Matthew Riedman
May 13, 2016

Using the direct comparison test, we see 1 + ( 1 2 ) + ( 1 3 + 1 4 ) + ( 1 5 + 1 6 + 1 7 + 1 8 ) . . . > 1 + ( 1 2 ) + ( 1 4 + 1 4 ) + ( 1 8 + 1 8 + 1 8 + 1 8 ) . . . = 1 + 1 2 + 1 2 + 1 2 . . . 1+\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)...>1+\left(\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)...=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}... which clearly diverges.

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Nice one, thanks!

Sandeep Bhardwaj - 5 years, 1 month ago
Sam Bealing
May 8, 2016

It is well known that the harmonic series does not converge and this can be shown by a comparison test.

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It would be great if you could explain the comparison test. Or give a proof of the harmonic series doesn't converge. Thanks :-)

Sandeep Bhardwaj - 5 years, 1 month ago

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We can look at it as (1\n)^p, as a p- series where p=1, thus it diverges.

Hana Wehbi - 5 years, 1 month ago

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