7

Prove that for all $n ≥ 1$ $$1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n} = \frac{1}{n+1} + \frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n}$$

My attempt:

By induction

Base case: $n = 1$

LHS : $1 - \frac{1}{2} = \frac{1}{2}$

RHS : $\frac{1}{1 + 1 } = \frac{1}{2}$

Suppose equality holds for $n$

We need to show that it holds for $n + 1$, i.e

$$1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}-+\cdots-\frac{1}{2n+2} = \frac{1}{n+2} + \frac{1}{n+3} +\frac{1}{n+4} +\cdots+\frac{1}{2n+2} $$

We have $$\begin{align} 1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\cdots-\frac{1}{2n+2} & = \bigl[1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots - \frac{1}{2n}\bigr] + \frac{1}{2n+1} - \frac{1}{2n+2} \\ & = \frac{1}{n+1} + \frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ & =\frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n} + \frac{1}{2n+1} +\bigl (\frac{1}{n+1} - \frac{1}{2n+2} \bigr)\\ & = \frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2} \end{align}$$

As desired. $\Box$

Is it correct?

José Carlos Santos
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Stokolos Ilya
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1 Answers1

4

Your induction works.

As an alternative,

  • writing $H_n =1 +\frac12 +\frac13 +\cdots +\frac1{n-1}+\frac1n$
  • then $H_{2n} =1 +\frac12 +\frac13 +\cdots +\frac1{2n-1}+\frac1{2n}$
  • while $\frac12H_n = \frac12 +\frac14 +\frac16 +\cdots +\frac1{2n-2}+\frac1{2n}$
  • so $H_{2n} - \frac12H_n =1 +\frac13 +\frac15 +\cdots +\frac1{2n-3} +\frac1{2n-1}$
  • and $H_{2n} - H_n =1 -\frac12 +\frac13 +\cdots +\frac1{2n-1}-\frac1{2n}$
  • which is equal to $H_{2n} - H_n = \frac{1}{n+1} +\frac{1}{n+2}+\frac{1}{n+3} +\cdots +\frac1{2n-1}+\frac1{2n}$

This then gives a reasonable approximation, since we know $H_n=\log_e{n}+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}-\cdots$ and so $H_{2n}=\log_e 2+\log_e{n}+\gamma+\frac{1}{4n}-\frac{1}{48n^2}+\frac{1}{1920n^4}-\cdots$ leading to $H_{2n}-H_n = \log_e 2 - \frac{1}{4n}+\frac{1}{16n^2} -\frac{1}{128n^4} +\cdots$

Henry
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