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Define $f: \mathbb{N} \rightarrow \mathbb{R}$ as $f(n)= \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}$.

I was wondering how to tell if $f$ is a increasing or decreasing function?

Thanks and regards!

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    @Tim: Then you take the last term $0$, that doesn't change much about the argument.2011-05-02

4 Answers 4

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A formal proof would be $ \begin{align} f(n+1) - f(n) &= \frac{1}{n+1} \sum_{i=1}^{n} \frac{1}{i} - \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i} \\ &= \frac{n}{n+1}\frac{1}{n} \left(\sum_{i=1}^{n-1} \frac{1}{i} + \frac{1}{n}\right) - \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i} \\ &= (\frac{n}{n+1} - 1)\frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i} + \frac{1}{n(n+1)} \\ & = \frac{1}{n(n+1)} \left(1 - \sum_{i=1}^{n-1} \frac{1}{i}\right) < 0 \end{align} $ for $\forall n \geq 3$

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    your welcome. +1 nice answer2011-05-02
7

A simpler proof would be to notice that $\displaystyle f(n) \gt \frac{1}{n}$ (for $n \gt 2$)

Thus $\displaystyle (n+1)f(n+1) - nf(n) = \frac{1}{n} \lt f(n)$ and so $f(n) \gt f(n+1)$.

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    @Tim/Jonas: Thanks! @Tim: I guess the intent is to have one answer which will make it easier for people who come across this later. We can always edit the accepted answer to have multiple proofs, I guess.2011-05-03
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The following 3 condidions are equivalent:

$f(n)>f(n+1)$

$\frac{1+\frac12+\dots+\frac1{n-1}}n>\frac{1+\frac12+\dots+\frac1{n}}{n+1}$

$n+\frac{n+1}2+\dots+\frac{n+1}{n-1}+1>n+\frac n2+\dots+\frac n{n-1}+1$

In the last inequality, the corresponding terms on the LHS are greater (or equal) as the corresponding terms on the RHS. (They both have the same number of terms.) At least one of these inequalities is strict.

EDIT: (From the comments I see that this was not clear enough.)

There is the same number of terms, since I divided $n+1$ (obtained by multypling the first term in the second inequality) between $n$ and $1$ (the first and the last term in the LHS).

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Consider: $\begin{align*} n(n+1)(f(n)-f(n+1))&=n(n+1)\left(\frac{1}{n}\sum_{i=1}^{n-1}\frac{1}{i}-\frac{1}{n+1}\sum_{i=1}^{n}\frac{1}{i}\right)\\ &=\sum_{i=1}^{n-1}\left(\frac{n+1}{i}-\frac{n}{i}\right) - 1\\ &=\sum_{i=1}^{n-1}\frac{1}{i} - 1\\ &\geq 0\end{align*}$ for $n\geq2$.

In specific $f(n)-f(n+1)\geq0$ in general.

Edit: Little writing error.

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    I do apologize, I forgot the term "-1". Now I edited the answer and it should be correct.2011-05-03