1)Base of induction n=2 : $${1+\frac{1}{2}+\frac{1}{3}>\frac{2}{2}}$$ 2)Assuming right for n=k $$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k-1} > \frac{k}{2}$$ and now need to prove that right for n=k+1 $$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k-1}+\frac{1}{2^k}+\frac{1}{2^k+1}+...+\frac{1}{2^{k+1}-1} > \frac{k+1}{2}$$ After simlifiying that we get to this $$\frac{1}{2^k}+\frac{1}{2^k+1}+...+\frac{1}{2^{k+1}-1}>\frac{1}{2}$$ any hints about further steps?
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The title and the question are different. – E. Joseph Oct 26 '16 at 15:26
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This formula is different to the one in your title. Can you write it in summation form so we can easily see the term by term pattern? – Hugh Oct 26 '16 at 15:27
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not true for $k=1$. – hamam_Abdallah Oct 26 '16 at 15:29
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Don't you need to show that $$\frac1{2^k}+\frac1{2^k+1}+...+\frac1{2^{k+1}\color{#C00000}{-}1}>\frac{1}{2}$$ instead? – robjohn Oct 26 '16 at 15:41
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@robjohn yeah, my mistake – KapfellFrau Oct 26 '16 at 15:42
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The previous inequality needs a similar fix – robjohn Oct 26 '16 at 15:43
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Hint: To show $$ \frac1{2^k}+\frac1{2^k+1}+...+\frac1{2^{k+1}-1}>\frac12 $$ count the number of terms and look at the smallest term.
robjohn
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so like $$\frac{1}{2^k}+\frac{1}{2^k+1}+...+\frac{1}{2^{k+1}-1}>\frac{2^k}{2^{k+1}-1}$$ and from there divide it by $2^k$ – KapfellFrau Oct 26 '16 at 15:48
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Well, $\ge$, but can you show that $\frac{2^k}{2^{k+1}-1}\gt\frac12$? Oh, you mean divide both numerator and denominator by $2^k$. Yeah. – robjohn Oct 26 '16 at 15:49
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if divide by $2^k$ i get $\frac{1}{2-\frac{1}{2^k}}$, as $2-\frac{1}{2^k} < 2 $ then it workes just fine i guess – KapfellFrau Oct 26 '16 at 15:53
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Perhaps simpler is to note that each of the $2^k$ terms is $\gt\frac1{2^{k+1}}$. – robjohn Oct 26 '16 at 15:53