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

[anemone]

Prove that $2^{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{n}\right)}_{\phantom{i}}<n$ for all integer $n\ge 2$.

 

[chisigma]

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

$\chi$ $\sigma$

 

[kaliprasad]

Spoiler

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

 

[MarkFL]

My solution:

Spoiler

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

\(\displaystyle 2^{\frac{1}{2}}<2\)

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

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

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

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

Now, consider that:

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

Hence:

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

Thus, we must have:

\(\displaystyle 2^{\dfrac{1}{n+1}}<1+\frac{1}{n}\)

or:

\(\displaystyle n2^{\dfrac{1}{n+1}}<n+1\)

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

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

or:

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

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