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Find Summation of n

Albert

Well-known member
Jan 25, 2013
1,225
$ n\in N$

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

find :$ \sum n $
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
My solution:

I would write the given inequality as:

\(\displaystyle n-n^{\frac{1}{2}}-n^{\frac{1}{3}}-n^{\frac{1}{4}}<0\)

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

\(\displaystyle n^{\frac{3}{4}}-n^{\frac{1}{4}}-n^{\frac{1}{12}}-1<0\)

Let \(\displaystyle u=n^{\frac{1}{12}}\) and we have:

\(\displaystyle u^9-u^3-u-1<0\)

Defining:

\(\displaystyle f(u)=u^9-u^3-u-1\)

we see that:

\(\displaystyle f(1)<0\) and \(\displaystyle f(2)>0\)

and:

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

we see also that:

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

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

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

Using the definition of $f$, we have:

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

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

\(\displaystyle u_1=1.4\)

\(\displaystyle u_2\approx1.27679115672466\)

\(\displaystyle u_3\approx1.19600432443480\)

\(\displaystyle u_4\approx1.16202903329198\)

\(\displaystyle u_5\approx1.15671666316563\)

\(\displaystyle u_6\approx1.15660010278918\)

\(\displaystyle u_7\approx1.15660004786155\)

\(\displaystyle u_8\approx1.15660004786153\)

\(\displaystyle u_9\approx1.15660004786153\)

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

\(\displaystyle n=u^{12}\approx5.73057856869580\)

Hence:

\(\displaystyle n\in\{1,2,3,4,5\}\)

And so:

\(\displaystyle \sum n=15\)
 

Albert

Well-known member
Jan 25, 2013
1,225
$ n\in N$

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

find :$ \sum n $

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