Prove U can't get an constant.

Let us make a series. S=1+12+13+14+15+16..............S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}.............. Prove that this will not converge to a constant.

Extra Credit: Do it without use of calculus.

Note by Utsav Singhal
4 years, 9 months ago

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This is popularly known as the divergence of the Harmonic Series

Agnishom Chattopadhyay Staff - 4 years, 9 months ago

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The proof that I like is by contradiction.

Suppose SS converges to a finite number. Then, show that S=S+1 S = S + 1 .

(details to follow).

Calvin Lin Staff - 4 years, 6 months ago

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It can be done using RATIO TEST.

Sandeep Bhardwaj - 4 years, 9 months ago

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Please give the details or any link from where I can learn it.. Thanks.

Satvik Golechha - 4 years, 9 months ago

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http://en.wikipedia.org/wiki/Ratio_test

Sandeep Bhardwaj - 4 years, 9 months ago

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@Sandeep Bhardwaj But the limit L here is 1. Therefore, test becomes inconclusive

Brilliant Member - 4 years, 6 months ago

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Let us make a AM HM inequality. 1/(a-1) + 1/a + 1/(a+1) >= 3/a So we will take its minnimum value... as 3/a Now lets put the values . Leaving 1 then taking 1/2 , 1/3, 1/4 then next three....then next three... So we will get its min value as 1 + 1 + 1/2 + 1/3. ..... to infinity. Then again doing it. 1+1+1+1+1+1+..................to infinite +1/2 + 1/3 + 1/4 .... So its minnimum value comes out to be infinite .

Soorry can't make the solution latex ...!

Utsav Singhal - 4 years, 8 months ago

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Even though the harmonic series does not converge, Is it possible for a partial sum of the series to be an integer?

Brian Charlesworth - 4 years, 6 months ago

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2.30 approx

Kanhaiya Yadav - 4 years, 9 months ago

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No no, it is totally wrong . There is no value of it . It will be infinite

Utsav Singhal - 4 years, 9 months ago

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