<s>[INST] For any value of $n\ge5$, the value of $1+ \frac{1}{2}+ \frac{1}{3}+...+\frac{1}{2^n+1}$ lies between 
For any value of $n\ge5$, the value of $$1+ \frac{1}{2}+ \frac{1}{3}+...+\frac{1}{2^n+1}$$ lies between
A) $0$ and $\frac{n}{2}$;
B) $\frac{n}{2}$ and $n$;
C) $n$ and $2n$;
D) none of them

I know this is Harmonic progression, but how exactly am I supposed to know the interval?
 [/INST] Note that
$$\sum\limits_{k=1}^{2^n+1} \dfrac{1}{k} > \sum\limits_{k=2}^{2^n}\dfrac{1}{k} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dots \dfrac{1}{2^n} > \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{8} + \dots + \dfrac{1}{2^n} = \sum\limits_{k=1}^n \dfrac{2^{k-1}}{2^k} = \dfrac{n}{2}$$
On the other hand,
$$\sum\limits_{k=1}^{2^n+1} \dfrac{1}{k} \le \sum\limits_{k=1}^{2^{n+1} - 1} \dfrac{1}{k} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dots \dfrac{1}{2^{n+1}-1} < 1+ \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dots + \dfrac{1}{2^n} = \sum\limits_{k=0}^{n} \dfrac{2^k}{2^k} = n + 1$$
You can decrease the upper bound to $n$ because $\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4}-\dfrac{1}{5} + \dfrac{1}{4}-\dfrac{1}{6} \dots + \dfrac{1}{2^{n+1}-1}-\dfrac{1}{2^n} > 1$.
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