<s>[INST] Can I show $\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$ by induction? It is known that $a_{n+1} > a_n$. I tried to prove
$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$$
via induction.
Base case: $a_0 = \frac{1}{0+1} = 1 \leq 1$
Inductive step: assume $a_n \leq 1$, prove $a_{n+1} \leq 1$.
$a_n = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$
$\frac{1}{n+2} + \frac{1}{n+3}+ \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2} \leq 1-\frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$
But $-\frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$ is positive, so I'm not sure where to go from here.
Normally, the inductive step would involve a change on the right hand side, but in this case, it remains the same. Makes me wonder if induction is even the correct approach?
 [/INST] According to Wolfram Alpha, this series is equivalent to $ \psi^{(0)}(2n+1)-\psi^{(0)}(n+1)$ where $ \psi^{(0)}(x)$ is the $n^{th}$ derivative of the digamma function. Using this, you could show the continuity for the function for values of $n > 0$, show that the limit as n approaches infinity is $log(2)$ (easily proven), and show that the function is continuously increases (by showing derivative is positive for $n>0$?)
Note that this actually shows that the series is always $< log(2)$, a stronger statement than that given by the induction argument.
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