Is it true that $1 - \frac 12+\frac13-\frac14+\cdots-\frac 1{200}=\frac 1{101}+\frac 1{102}+\cdots+\frac 1{200}$? Where can I find sources for this proof?
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1For a boring proof, try induction. Go from $2$ to $4$ to $6$ and so on. – Yuval Filmus Oct 09 '14 at 04:42
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@YuvalFilmus : induction? It does not work.. i tried that... – Oct 09 '14 at 04:43
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@YuvalFilmus : how can you edit the question if it also changes the meaning of the question... – Oct 09 '14 at 04:49
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@PraphullaKoushik I have lots of reputation and so the privilege to edit posts at will. The OP can always contest my edits, and if I am found to have done something wrong, the moderators can sanction me. In this case, there was an obvious mistake in the formula. I used my judgement to decide that rather than answering "no", I would correct the typo. – Yuval Filmus Oct 09 '14 at 04:51
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Hi Praphulla Koushik, Yuval Filmus was correct, I made a typo and started at $\frac 1{100}$ instead of $\frac 1{101}$. – someone Oct 09 '14 at 04:53
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@YuvalFilmus : You can ofcourse edit the question as you have lots of reputation but the question was totally different than your edit.. that is what my point is... I was not so sure that is an obvious mistake or not... – Oct 09 '14 at 04:53
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@someone : I know that he is correct.. I was saying you are the one who is supposed to edit the mistakes... – Oct 09 '14 at 04:56
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@YuvalFilmus : The comment that i have made was supposed to be noted for OP... I did {@}your name by mistake.. – Oct 09 '14 at 05:02
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@PraphullaKoushik I am a new user here, how can I edit my post? I thought only users with lots of reputation can do it, as Yuval Filmus wrote. – someone Oct 09 '14 at 05:07
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@someone : YOu can not edit other's post but you can definitely edit yours... – Oct 09 '14 at 05:48
4 Answers
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Hint: Add $$2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{200}\right)$$to both sides.
André Nicolas
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$$\sum_{r=n+1}^{2n}\frac1r=\sum_{r=1}^{2n}\frac1r-\sum_{r=1}^n\frac1r$$
$$=\sum_{r=1}^n\frac1{2r-1}+\sum_{r=1}^n\frac1{2r}-\sum_{r=1}^n\frac1r$$
Now $$\sum_{r=1}^n\frac1{2r}-\sum_{r=1}^n\frac1r=\sum_{r=1}^n\left(\frac1{2r}-\frac1r\right)=-\sum_{r=1}^n\frac1{2r}$$
lab bhattacharjee
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1@PraphullaKoushik, I reached at$$\sum_{r=n+1}^{2n}\frac1r=\sum_{r=1}^n\frac1{2r-1}-\sum_{r=1}^n\frac1{2r}$$ – lab bhattacharjee Oct 09 '14 at 04:53
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Here is the boring proof by induction. We want to prove that $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} = \frac{1}{n+1} + \cdots + \frac{1}{2n}. $$ Base case: $n = 1$. The left-hand side reads $1-1/2 = 1/2$, and the right-hand side is $1/2$.
Induction step: Suppose $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} = \frac{1}{n+1} + \cdots + \frac{1}{2n}. $$ Then $$ \begin{align*} &1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n+1} - \frac{1}{2n+2} \\ = &1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ = &\frac{1}{n+1} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ = &\frac{1}{n+2} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{n+1} - \frac{1}{2n+2} \\ = &\frac{1}{n+2} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}. \end{align*} $$
Yuval Filmus
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To make this a bit easier to think about... $$\sum\limits_{r=1}^{200} \frac{1}{r} = 1 + \frac{1}{2} + \frac{1}{3}+ \cdots + \frac{1}{199} + \frac{1}{200}$$
Subtract the even fractions between 1/2 and 1/200: $$\sum\limits_{r=1}^{200} \frac{1}{r} - \sum\limits_{r=1}^{100} \frac{1}{2r}= 1 + \frac{1}{3} + \cdots + \frac{1}{199}$$
Take the even fractions off again $$\sum\limits_{r=1}^{200} \frac{1}{r} - 2\sum\limits_{r=1}^{100} \frac{1}{2r}= 1 -\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{199} - \frac{1}{200}$$
Now, $$2\sum\limits_{r=1}^{100} \frac{1}{2r} = \sum\limits_{r=1}^{100} \frac{1}{r}$$
So the left-hand side becomes:
$$\sum\limits_{r=1}^{200} \frac{1}{r} - \sum\limits_{r=1}^{100} \frac{1}{r}$$ $$ = \sum\limits_{r=101}^{200} \frac{1}{r} = \frac{1}{101} + \frac{1}{102} + \cdots + \frac{1}{200} $$ (quod erat demonstrandum)
user23614
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