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$1-1/2+1/3-1/4+ \cdots +1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+ \cdots +1/2n$

I was asked to prove by mathematical induction the validity of the above equation. It isn't hard to prove that it holds for any arbitrary natural number. But how mathematicians (or anyone) discovered that the left side of that equation equals to the right side, it doesn't seem obvious. I've tried to manipulate them with various means such as multiplying the denominator but I can't observe any pattern. Is it by chance that this equation was discovered? Thanks in advance.

Dave Clifford
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1 Answers1

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I don't know how someone first discovered this identity, but here's a clearer way of seeing it: \begin{align*} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{2n} &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n}- 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} \right)\\ &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n}-\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \right) \end{align*}

This cancels out the first $n$ terms of the sequence, leaving the $(n+1)^\text{st}$ to $2n^\text{th}$ terms, which is the righthand side.

Viktor Vaughn
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    Thanks! But how do you know to introduce those numbers into it? Is it by experience? Can you introduce any book related to this topic? – Dave Clifford Feb 13 '14 at 06:41
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    I just thought about the difference between the sum with all positive terms, and the sum with the alternating sign. This difference happens to be 2 times the sum of the terms with even denominators. I don't know of a book dedicated to this topic, but here's a fun exercise that can be solved using the same trick: Given $p>1$, simplify the expression $$\frac{1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots}{1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots}.$$ (This is a Problems Plus exercise in Chapter 12 of Stewart's Calculus.) – Viktor Vaughn Feb 13 '14 at 14:56
  • @RichardD.James, what is the answer to the above problem, I mean answer not solution – Arjun May 15 '20 at 14:39
  • @ArjunRana The answer to the problem in my comment? Just consider $\left(1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots\right) - \left(1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots\right)$. In the end I get $\frac{2^{p-1}}{2^{p-1}-1}$. – Viktor Vaughn May 15 '20 at 15:09
  • @Richard D. James, thanks, I got the same answer. Wanted to get it verified – Arjun May 15 '20 at 15:11