Can You Determine the Sum of All Natural Numbers Less Than Their Combined Roots?

[Albert1]

$ n\in N$

$n<\sqrt n + \sqrt[3]{n} + \sqrt[4]{n}$

find :$ \sum n $

 

[MarkFL]

My solution:

Spoiler

I would write the given inequality as:

$$n-n^{\frac{1}{2}}-n^{\frac{1}{3}}-n^{\frac{1}{4}}<0$$

Divide through by $$n^{\frac{1}{4}}>0$$ to obtain:

$$n^{\frac{3}{4}}-n^{\frac{1}{4}}-n^{\frac{1}{12}}-1<0$$

Let $$u=n^{\frac{1}{12}}$$ and we have:

$$u^9-u^3-u-1<0$$

Defining:

$$f(u)=u^9-u^3-u-1$$

we see that:

$$f(1)<0$$ and $$f(2)>0$$

and:

$$f'(u)=9u^8-3u^2-1$$

we see also that:

$$f'(x)>0$$ for $$1\le x$$

So we may apply Newton's method to find the real root of $f$ on $(1,2)$.

$$u_{n+1}=u_n-\frac{f\left(u_n \right)}{f'\left(u_n \right)}$$

Using the definition of $f$, we have:

$$u_{n+1}=u_n-\frac{u_n^9-u_n^3-u_n-1}{9u_n^8-3u_n^2-1}=\frac{8u_n^9-2u_n^3+1}{9u_n^8-3u_n^2-1}$$

Letting $u_0=1$, we then recursively obtain:

$$u_1=1.4$$

$$u_2\approx1.27679115672466$$

$$u_3\approx1.19600432443480$$

$$u_4\approx1.16202903329198$$

$$u_5\approx1.15671666316563$$

$$u_6\approx1.15660010278918$$

$$u_7\approx1.15660004786155$$

$$u_8\approx1.15660004786153$$

$$u_9\approx1.15660004786153$$

Hence we know $$u\approx1.15660004786153$$ is the only real root of $f$ on $[1,\infty)$ and so:

$$n=u^{12}\approx5.73057856869580$$

Hence:

$$n\in\{1,2,3,4,5\}$$

And so:

$$\sum n=15$$

 

[Albert1]

$ n\in N$

$n<\sqrt n + \sqrt[3]{n} + \sqrt[4]{n}$

find :$ \sum n $

Spoiler

$n<\sqrt n + \sqrt[3]{n} + \sqrt[4]{n}=k$
$3\sqrt[3]{n} <k<3 \sqrt[2]{n}$
if $n<3\sqrt[2]{n}$,then $n<9---(1)$
if $n<3\sqrt[3]{n}$,then $n<6---(2)$
for :$7=3+2+2=\sqrt 9+\sqrt [3]{8} +\sqrt[4]{16}>\sqrt 7+\sqrt[3]{7}+\sqrt[4]{7}$
$\therefore \sum n =1+2+3+4+5=15$