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}..............S=1+21+31+41+51+61.............. Prove that this will not converge to a constant.
Extra Credit: Do it without use of calculus.
Note by Utsav Singhal 6 years, 8 months ago
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2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
This is popularly known as the divergence of the Harmonic Series
The proof that I like is by contradiction.
Suppose SSS converges to a finite number. Then, show that S=S+1 S = S + 1 S=S+1.
(details to follow).
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.
http://en.wikipedia.org/wiki/Ratio_test
@Sandeep Bhardwaj – But the limit L here is 1. Therefore, test becomes inconclusive
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 ...!
Even though the harmonic series does not converge, Is it possible for a partial sum of the series to be an integer?
2.30 approx
No no, it is totally wrong . There is no value of it . It will be infinite
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Easy Math Editor
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.
When posting on Brilliant:
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or__bold__
paragraph 1
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> This is a quote
\(
...\)
or\[
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
This is popularly known as the divergence of the Harmonic Series
The proof that I like is by contradiction.
Suppose S converges to a finite number. Then, show that S=S+1.
(details to follow).
It can be done using RATIO TEST.
Log in to reply
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
Log in to reply
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 ...!
Even though the harmonic series does not converge, Is it possible for a partial sum of the series to be an integer?
2.30 approx
Log in to reply
No no, it is totally wrong . There is no value of it . It will be infinite