Proving the Harmonic Series Sum Formula for Positive Integers | Math Proof

AI Thread Summary
The discussion revolves around proving the formula for the sum of harmonic series for positive integers, specifically that the sum of harmonic numbers up to n equals (n+1)H_n - n. The user calculates H_5 but finds a discrepancy when applying the formula. Another participant clarifies that the correct approach involves summing H_1 through H_5, which aligns with the expected result. The conversation highlights the importance of correctly interpreting the series and applying the formula accurately. Understanding the harmonic series and its properties is crucial for resolving such mathematical proofs.
James889
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Hai,

The harmonic series is given by: H_{n} = \sum_{i=1}^n \frac{1}{i}

I need to prove that for all positive integers:
\sum_{j=1}^n H_{j} = (n+1)H_{n} -n

So i have
H_{5} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{137}{60}

H_{5} \neq (5+1)*\frac{137}{60} -5

Have i missed something here?

Please excuse my epic fail math skills...
 
Last edited:
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So for your H5 example what they want you to sum is

H1 + H2 + H3 + H4 + H5

When I did that I got what the problem tells you you will get
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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