Show that $\frac{1}{3}+\frac{8}{7}+\cdots +\frac{n^3}{n^2+n+1}<\frac{n(3n+5)}{6}$

Question

Show that $$\frac{1}{3}+\frac{8}{7}+\frac{27}{13}+\cdots +\frac{n^3}{n^2+n+1}<\frac{n(3n+5)}{6}$$

for all natural number $\;\forall\; n\geq 1$

what i try

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

$$\frac{n^3}{n^2+n+1}<\sum^{n}_{k=1}n^2-n+1=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+n=\frac{n(n^2+2)}{3}$$

How do i solve it Help me please

Answer 1

Induction will do; you need to show \begin{eqnarray*} \frac{n(3n+5)}{6}+ \frac{(n+1)^3}{(n+1)^2+n+1+1} < \frac{(n+1)(3n+8)}{6}. \end{eqnarray*} This is the same as $0<8n^2+24n+18$, which is easy to show.

Answer 2

You are considering $$\sum_{k=1}^n \frac{k^3}{k^2+k+1}=\sum_{k=1}^n (k-1)+\sum_{k=1}^n \frac{1}{k^2+k+1}=\frac{n(n-1)}2+\sum_{k=1}^n \frac{1}{k^2+k+1}$$ $$k^2+k+1 > k^2 \implies \sum_{k=1}^n \frac{1}{k^2+k+1}<\sum_{k=1}^n \frac{1}{k^2}<\sum_{k=1}^\infty \frac{1}{k^2}=\frac {\pi^2}6$$ So, since we have now an upper bound of the lhs, it is sufficient to prove that $$\frac{n(n-1)}2+\frac {\pi^2}6<\frac{n(3n+5)}{6}\implies \frac {\pi^2}6<\frac{4n}3$$ which is true $\forall n >0$

Answer 3

$$\sum_{k=1}^n\frac{k^3}{k^2+k+1}<\sum_{k=1}^nk=\frac{n(n+1)}{2}<\frac{n(3n+5)}{6}$$

Answer 4

Easy induction.- True for $n=1$ and $$\frac{1}{3}+\frac{8}{7}+\frac{27}{13}+\cdots +\frac{n^3}{n^2+n+1}<\frac{n(3n+5)}{6}$$ $$\frac{1}{3}+\frac{8}{7}+\frac{27}{13}+\cdots +\frac{n^3}{n^2+n+1}+\frac{(n+1)^3}{(n+1)^2+(n+1)+1}<\frac{(n+1)(3(n+1)+5)}{6}$$ Besides $$\frac{(n+1)(3(n+1)+5)}{6}=\frac{n(3n+5)}{6}+\frac{6n+4}{6}$$ Then by hypothesis of induction we have to prove that $\dfrac{(n+1)^3}{(n+1)^2+(n+1)+1}\le\dfrac{6n+4}{6}$ which reduces to $$0\lt2n+1$$