Krdytkyu

Prove that for every $n$ we have $$sum_{k=1}^n frac{1}{k^4} le 2 - dfrac{1}{sqrt{n}}$$ I've tried induction, but I ended up with polynomials of high degree.

For $nge2$,
$$
begin{align}
frac1{n^4}
&lefrac1{2n^{3/2}}\
&=frac1{sqrt{nvphantom{-1}}sqrt{nvphantom{-1}}left(sqrt{nvphantom{-1}}+sqrt{nvphantom{-1}}right)}\
&lefrac1{sqrt{nvphantom{-1}}sqrt{n-1}left(sqrt{nvphantom{-1}}+sqrt{n-1}right)}\
&=frac1{sqrt{n-1}}-frac1{sqrt{nvphantom{-1}}}
end{align}
$$
Therefore,
$$
begin{align}
sum_{k=1}^nfrac1{k^4}
&le1+sum_{k=2}^nleft(frac1{sqrt{k-1}}-frac1{sqrt{kvphantom{-1}}}right)\
&=2-frac1{sqrt{nvphantom{-1}}}
end{align}
$$

The function $x^{-p}$ is a positive decreasing function. For such functions, sums at evenly spaced points are well approximated by integrals. More precisely,$$ int_1^n frac{1}{x^p}dx < sum_{i = 1}^n frac{1}{i^p} < int_1^n
frac{1}{x^p}dx + 1.$$

$$sum_{k=1}^n frac{1}{k^4}<int_1^n
frac{1}{x^4}dx + 1 =1+frac{n^{-3}}{-3}=1-frac{1}{3n^3}$$
Also, given that the function $$f(x)=3x^3sqrt{x}+sqrt{x}-3x^3$$
is strictly increasing and defined on $[0,+infty)$ with $f(0)=0$, we have

$$f(n) geq 0 iff 3n^3sqrt{n}+sqrt{n}-3n^3geq0 Rightarrow frac{3n^3sqrt{n}+sqrt{n}-3n^3}{3n^3sqrt{n}}> 0 iff$$
$$ 1+frac{1}{3n^3}-frac{1}{sqrt{n}} >0 Rightarrow1-frac{1}{3n^3}< 2-frac{1}{sqrt{n}}$$

So
$$sum_{k=1}^n frac{1}{k^4}leq2-frac{1}{sqrt{n}}$$
with equality only for $n=1$.