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I'm trying to show that $$\sum_{i=1}^n ln(1+\frac{1}{i}) \leq \sum_{i=1}^{n}\frac{1}{i}$$ for all $n\in\mathbb{N}$. My first instinct is to go the route of saying that it is sufficient to show that $$\ln(1+\frac{1}{i}) \leq \frac{1}{i}$$ for all $i\in\mathbb{N}$. So first off, is this valid (as an axiom, or the result of a basic enough proof that I don't need to prove this statement)?

Assuming that that's OK, my next challenge is to prove the latter statement. It's obvious on its face, but that's not rigorous. My instinct then is to use an initial value + derivative:

  1. show that the left side is less than the right side for $i=1$
  2. show that the derivative of the left side is less than the derivative of the right side
  3. therefore the left side must always be less than the right side.

I'm not sure if I can use this though for two reasons - firstly, because I'm not sure what this theorem is called (the initial value + derivative theorem?) and I'm not sure it's basic enough to use without a proof. And secondly, I'm not sure you can use calculus, which is continuous, to prove properties of discrete numbers. So, maybe I'm off here. Am I alright on this track, or if not, does anyone see a better way to go about proving this?

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One big tip: the entire left hand side collapses. $\ln(a) + \ln(b) = \ln(ab)$, and consecutive terms are e.g. $1+1/i = \frac{i+1}{i}$ and $1+1/(i+1) = \frac{i+2}{i+1}$, etc; this lets you get a much simpler form for the LHS as a whole, and from there you should be able to use known results on the harmonic series... – Steven Stadnicki Oct 9 '12 at 21:36

If you prove $\log(1+(1/i))\le1/i$ for all real $i\ge1$, then you have, a fortiori, proved it for all positive integers $i$.

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To address your first question, if $a \le b$ and $c \le d$, then $a+c\le b+c \le b + d $. You can see that in general it is always okay to add two inequalities. Also, if you want to prove that the sum of one series is greater than the sum of another series, it is sufficient to show that each term of the first series is greater than that of the second.

It is easy to prove that $\log(1+x)\le x$ for all $x>0$. Consider $f(x) = x - \log(1+x)$. $f(0)=0$ and $f'(x)=1-1/(1+x)\ge 0 $ for all $x\ge0$.

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