Let us make a series. \(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}..............\) Prove that this will not converge to a constant.

**Extra Credit:** Do it without use of calculus.

Let us make a series. \(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}..............\) Prove that this will not converge to a constant.

**Extra Credit:** Do it without use of calculus.

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TopNewestThis is popularly known as the divergence of the Harmonic Series

– Agnishom Chattopadhyay · 2 years, 4 months agoLog in to reply

The proof that I like is by contradiction.

Suppose \(S\) converges to a finite number. Then, show that \( S = S + 1 \).

(details to follow). – Calvin Lin Staff · 2 years, 1 month 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 · 2 years, 1 month 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 · 2 years, 4 months ago

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It can be done using RATIO TEST. – Sandeep Bhardwaj · 2 years, 4 months ago

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– Satvik Golechha · 2 years, 4 months ago

Please give the details or any link from where I can learn it.. Thanks.Log in to reply

http://en.wikipedia.org/wiki/Ratio_test – Sandeep Bhardwaj · 2 years, 4 months ago

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– Subrata Saha · 2 years, 1 month ago

But the limit L here is 1. Therefore, test becomes inconclusiveLog in to reply

2.30 approx – Kanhaiya Yadav · 2 years, 4 months ago

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– Utsav Singhal · 2 years, 4 months ago

No no, it is totally wrong . There is no value of it . It will be infiniteLog in to reply