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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 .

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestThis is popularly known as the divergence of the Harmonic Series

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The proof that I like is by contradiction.

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

(details to follow).

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It can be done using RATIO TEST.

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

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

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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 ...!

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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?

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2.30 approx

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

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