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I'm learning calculus, specifically limit of sequences and derivatives, and need help with the following exercise:

Show that for every $n > 1$,

$$\log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n - 1} \quad \quad (1)$$

Important: this exercise is the continuation of a previous problem showing that, for every $n > 1$,

$$\frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n \quad \quad (2)$$

A detailed solution of the latter inequality using the MVT can be found here.

Now back to inequality $(1)$. My first guess was to use mathematical induction in order to prove it but I didn't get far. I think I should make use of inequality $(2)$ from the previous exercise but I'm stuck here.

Community
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glpsx
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  • Apply n through 2n to the right equality from (2), and you get n+1 inequalities. Add the two sides of these inequalities and you will get the left inequality. In a similar way, you can prove the right inequality. – Huang Aug 18 '16 at 00:13
  • Incidentally this gives an easy proof that $1/n + 1/(n + 1) + \cdots + 1/2n$ tends to $\log 2$ as $n \to \infty$. – Paramanand Singh Aug 18 '16 at 08:07

3 Answers3

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Hint:

To prove the left half of the inequality, start by noting that $$ \begin{align*} \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}&>\log\left(1+\frac{1}{n}\right)+\log\left(1+\frac{1}{n+1}\right)+\cdots\log\left(1+\frac{1}{2n}\right)\\ &=\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n+1}\right)+\cdots+\log\left(\frac{2n+1}{2n}\right). \end{align*} $$ How can you simplify this sum of logarithms?

Nick Peterson
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  • Using the identity $\log a + \log b = \log ab$, if we successively add every pair together, factors inside the logarithm will simplify. For instance, if we add the first two terms together, the $(n + 1)$ factors will cancel. At the end of the process we are left with the $(2n + 1)$ factor from the last term but I don't get how the denominator simplifies to $n$. Is that the correct idea? – glpsx Aug 18 '16 at 00:53
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    You've got it! It doesn't "simplify" to $n$ -- $n$ has been the denominator from the beginning! At each combination, you are cancelling the current numerator with the next denominator. So, at the end all that survives is the first denominator ($n$), and the last numerator ($2n+1$). – Nick Peterson Aug 18 '16 at 04:20
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For the LHS it is $$\frac { 1 }{ n } +\frac { 1 }{ n+1 } +\cdots +\frac { 1 }{ 2n } =\int _{ n }^{ n+1 }{ \frac { 1 }{ n } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ n+2 } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ 2n } dx } >\\ >\int _{ n }^{ n+1 }{ \frac { 1 }{ x } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ x } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ x } dx } =\int _{ n }^{ 2n+1 }{ \frac { 1 }{ x } } dx=\log \frac { 2n+1 }{ n } $$ and for the RHS it is $$\frac { 1 }{ n } +\frac { 1 }{ n+1 } +\cdots +\frac { 1 }{ 2n } =\int _{ n }^{ n+1 }{ \frac { 1 }{ n } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ n+2 } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ 2n } dx } <\\ <\int _{ n-1 }^{ n }{ \frac { 1 }{ x } dx } +\int _{ n }^{ n+1 }{ \frac { 1 }{ x } dx } +...+\int _{ 2n-1 }^{ 2n }{ \frac { 1 }{ x } dx } =\int _{ 2n-1 }^{ 2n }{ \frac { 1 }{ x } } dx=\log \frac { 2n }{ n-1 } \\ $$

haqnatural
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For the sake of completeness, let we show how to get a tight upper bound with the same technique used by Nick Peterson for the lower bound.

Since for any $|x|<1$ we have $$ \text{arctanh}(x) = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right) = x + \frac{x^3}{3}+\frac{x^5}{5}+\ldots \tag{1} $$ by evaluating $(1)$ at $x=\frac{1}{n}$ we get: $$ \sum_{k=n}^{2n}\frac{1}{k} < \frac{1}{2}\sum_{k=n}^{2n}\log\left(\frac{k+1}{k-1}\right)=\frac{1}{2}\log\prod_{k=n}^{2n}\frac{k+1}{k-1}=\frac{1}{2}\log\left(\frac{4n+2}{n-1}\right). \tag{2}$$

Jack D'Aurizio
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