Prove that $\frac{1}{1^4}+\frac{1}{2^4}+\cdots+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$

Question

I have to show: $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ for natural $n$

I tried to show it by induction (but I think it could be possible to show it using some ineqaulity of means) so for $n=1$ we have $1=1$ so inequality holds then I assume it's true for $n$ and for $n+1$ my thesis is $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n+1}}$

I know that: $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}$ but later I'm not sure if I have to show

$\displaystyle2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}\le2-\frac{1}{\sqrt{n+1}}$ or should be $\ge$

Answer 1

You want to show that $$2-\frac{1}{\sqrt{n}}+\frac1{(n+1)^4}\le 2-\frac1{\sqrt{n+1}}.\tag{1}$$ If (1) is true, then using induction hypothesis one gets $$1+\frac1{2^4}+\cdots+\frac{1}{n^4}+\frac{1}{(n+1)^4}\le 2-\frac1{\sqrt{n}}+\frac1{(n+1)^4}\le 2-\frac1{\sqrt{n+1}},$$ and we are done by induction.

Showing (1) should be easy since $$\frac{1}{\sqrt{n}}-\frac1{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac1{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})}.$$

Answer 2

You should show its:

$\displaystyle2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}\le2-\frac{1}{\sqrt{n+1}}$

The other ways does not help your proof (think about it for a while)

Answer 3

If we are allowed to use tools from the calculus, we can show that the result holds at $n=1,2$, and then use the fact that $$\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\cdots\lt \int_2^\infty x^{-4}\,dx.$$

From this we can conclude that our sum $S_n$ is always less than $1+\frac{1}{16}+\frac{1}{24}$.