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$$\sum_{i=1}^n \frac{1}{n+i} = \sum_{i=1}^{2n} \frac{(-1)^{i+1}}{i}$$

$$\prod_{i=1}^n (1-x_i) \gt 1-\sum_{i=1}^{n} x_i$$

Can someone please help me with these two? For the first one I have never seen two variables in one induction question and for the second one I have never done anything with sigma notation, so don't know how to proceed...

Thanks!!

Jerry Cohen
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1 Answers1

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Just change the variable. Instead of summing over $i$, sum over some variable $j=n+i$. Then the first value is going to be $n+1$ and the last value is $n+n=2n$. So you write $$\sum_{i=1}^n\frac1{n+i}=\sum_{j=n+1}^{2n}\frac1j$$ For the second part neither the product terms or the sum terms depend on $i$. So you rewrite that as $$(1-x)^n>1-nx$$ EDIT

So let's see what that means for part 1: take $n=1$. The left hand side is $$\sum_{i=1}^1\frac{1}{1+i}=\frac12$$ The right hand side is $$\sum_{i=1}^2\frac{(-1)^{i+1}}{i}=\frac{(-1)^2}{1}+\frac{(-1)^3}{2}=1-\frac12=\frac12$$ Let's try $n=2$. Left hand side is $$\sum_{i=1}^2\frac1{2+i}=\frac13+\frac14=\frac7{12}$$ Notice that I wrote this as $$\sum_{j=2+1=3}^{2\cdot2=4}\frac 1j$$ The right hand side is $$\sum_{i=1}^{2\cdot 2=4}\frac{(-1)^{i+1}}i=1-\frac12+\frac13-\frac14=\frac{12-6+4-3}{12}=\frac7{12}$$ So you equality holds for $n=1$ and $n=2$. The induction method says that if the statement holds for $n$, then it should hold for $n+1$ as well. So we know that $$\sum_{i=1}^n\frac1{n+i}=\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i}$$ We want to show that$$\sum_{i=1}^{n+1}\frac1{n+1+i}=\sum_{i=1}^{2(n+1)}\frac{(-1)^{i+1}}{i}$$

Now I'm going to use my hint and write the left hand side as $$\sum_{j=n+2}^{2n+2}\frac1j=\sum_{j=n+1}^{2n+2}\frac1j-\frac1{n+1}=\sum_{j=n+1}^{2n}\frac1j-\frac1{n+1}+\frac1{2n+1}+\frac1{2n+2}$$ Notice that I added a term at the beginning, which I had to subtract, and the last two terms I wrote them explicitly. Why this way? Because now the sum is the same as for the base case. $$\sum_{j=n+1}^{2n}\frac1j-\frac1{n+1}+\frac1{2n+1}+\frac1{2n+2}=\sum_{i=1}^{2n}\frac{(-1)^{i+1}}i+\frac1{2n+1}-\frac1{2n+2}\\=\sum_{i=1}^{2n}\frac{(-1)^{i+1}}i+\frac{(-1)^{2n+2}}{2n+1}+\frac{(-1)^{2n+3}}{2n+2}=\sum_{i=1}^{2n+2}\frac{(-1)^{i+1}}i$$

Andrei
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  • To the proposer: The 2nd Q is valid if we assume $0\le x\le 1$. – DanielWainfleet Sep 29 '21 at 00:39
  • ehmmm for the first one: i still dont get it... sorry could you explaine abit more – Jerry Cohen Sep 29 '21 at 05:49
  • and the second one, i wrote the question wrong sorry.... – Jerry Cohen Sep 29 '21 at 05:50
  • after 5 hours on it i still dont know how to proceed.... – Jerry Cohen Sep 29 '21 at 09:48
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    @JerryCohen I am not sure where you have problems with the first part. Did you perform the change of variable as I've mentioned? – Andrei Sep 29 '21 at 14:12
  • I just dont know how to even start with two different variables... for example does n stay a constant and doesnt need to be summed up? and i would be the only thing to sum up? and i dont quite understand how to do the induction when you have "2n" instead of "n"...also you used j instead of what was before, doesnt that change the whole equation? Sorry a lot of questions ... – Jerry Cohen Sep 29 '21 at 16:31
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    See if you can understand now the first part – Andrei Sep 29 '21 at 18:18
  • @Andrei I cannot believe smart people like you are willing to spend your time to help strangers from the internet. Everything is so clear now! Thank you so much. Words cannot express my feeling now!!!!!!!!!!!!Thank you thank you!!!!!!!!!!!!!!!!!! – Jerry Cohen Sep 29 '21 at 19:45
  • ...i hate to ask but can you also give me a hint on the second one? never done a product induction before.... – Jerry Cohen Sep 30 '21 at 05:03
  • See cases when $n=1,2$. Then assume is true for $n$. Multiply both sides with $(1-x_{n+1})$. – Andrei Sep 30 '21 at 12:16