MHB Proving $2^{\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}}<n$ for All $n\ge 2$

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The discussion focuses on proving the inequality \(2^{\left(\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}\right)} < n\) for all integers \(n \ge 2\). It references the asymptotic behavior of the harmonic series, noting that the limit of the difference between the harmonic sum and the natural logarithm approaches Euler's constant, \(\gamma\). As \(n\) increases, the logarithmic transformation shows that the left side grows slower than the right side, confirming the inequality holds for sufficiently large \(n\). The proof relies on the properties of logarithms and the behavior of the harmonic series. The conclusion reinforces that the inequality is valid for all integers \(n\) starting from 2.
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Prove that $2^{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{n}\right)}_{\phantom{i}}<n$ for all integer $n\ge 2$.
 
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anemone said:
Prove that $2^{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{n}\right)}_{\phantom{i}}<n$ for all integer $n\ge 2$.

[sp]Is...

$\displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k} - \ln n = \gamma\ (1)$

... where $\gamma = .5772... $ is thye Euler's constant, so that...

$\displaystyle \lim_{n \rightarrow \infty} \sum_{k=2}^{n} \frac{1}{k} - \ln n = \gamma - 1 < 0\ (2)$

... and that means that for n 'large enough' ...

$\displaystyle \ln \{ 2^{(\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n})} \} < \ln \{ e^{(\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n})} \} < \ln n\ (3)$ [/sp]

Kind regards

$\chi$ $\sigma$
 
we have
$(\dfrac{k}{k-1})^k = (1 + \dfrac{1}{k-1})^k \ge 1 + k \dfrac{1}{k-1} \gt 2$
or $2^\dfrac{1}{k} < (\dfrac{k}{k-1})$
multiply n- 1 terms taking k from 2 to n we get the result
 
My solution:

We see that for the base case $n=2$, we have:

$$2^{\frac{1}{2}}<2$$

This is true, so we may state the induction hypothesis $P_k$:

$$2^{\sum\limits_{j=2}^k\left(\dfrac{1}{j}\right)}<k$$

If, as our induction step, we multiply $P_k$ by $$2^{\dfrac{1}{k+1}}$$, there results:

$$2^{\sum\limits_{j=2}^{k+1}\left(\dfrac{1}{j}\right)}<k\cdot2^{\dfrac{1}{k+1}}$$

Now, consider that:

$$0<\sum_{k=2}^{n+1}\left({n+1 \choose k}\frac{1}{n^k}\right)$$

Hence:

$$2<2+\frac{1}{n}+\sum_{k=2}^{n+1}\left({n+1 \choose k}\frac{1}{n^k}\right)=\sum_{k=0}^{n+1}\left({n+1 \choose k}\frac{1}{n^k}\right)=\left(1+\frac{1}{n}\right)^{n+1}$$

Thus, we must have:

$$2^{\dfrac{1}{n+1}}<1+\frac{1}{n}$$

or:

$$n2^{\dfrac{1}{n+1}}<n+1$$

This means, going back to our induction, we may now state:

$$2^{\sum\limits_{j=2}^{k+1}\left(\dfrac{1}{j}\right)}<k\cdot2^{\dfrac{1}{k+1}}<k+1$$

or:

$$2^{\sum\limits_{j=2}^{k+1}\left(\dfrac{1}{j}\right)}<k+1$$

We have derived $P_{k+1}$ from $P_k$, thereby completing the proof by induction.
 
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