gouthamphilomath wrote: $(HUN 2)$ Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}<\frac{5}{4}$ $1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{5^2}=1.463611...$ $1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots=\frac{\pi^2}{6}<1.66$ Hence $\frac{1}{6^3}+\frac{1}{7^3}+frac{1}{8^3}+\frac{1}{9^3}+\cdots<\frac{\frac{1}{6^2}+\frac{1}{7^2}+frac{1}{8^2}+\frac{1}{9^2}+\cdots}{6}<\frac{1}{30}$ So we find $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}<\frac{49}{40}$ and hence result.

gouthamphilomath wrote: $(HUN 2)$ Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}<\frac{5}{4}$ Since $n^3 >(n-1)n(n+1)$, and $\frac{1}{(n-1)\cdot n \cdot (n+1)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1} \right)$, we have: $1+\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{n^3}< 1+\frac{1}{1\cdot2 \cdot 3}+...+\frac{1}{(n-1)\cdot n \cdot (n+1)}$ $=1+ \frac{1}{2} \left( (1-\frac{2}{2}+\frac{1}{3} ) + (\frac{1}{2}-\frac{2}{3}+\frac{1}{4})+...+(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}) \right)$ $=1+\frac{1}{2} \left( \frac{1}{2}-\frac{1}{n}+\frac{1}{n+1} \right) < 1+ \frac{1}{4}=\frac{5}{4}$

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Let $a_n=\sum_{i=1}^n\frac1{i^3}$. In fact, we claim that $$a_n=\sum_{i=1}^n\frac1{i^3}\le\frac54-\frac1{4n}$$for all $n\in\mathbb N$, which is obviously stronger. The base case is $a_1\le\frac54-\frac14$, so now assume that the assertion holds for some $n$. Then: \begin{align*}a_{n+1}&=\sum_{i=1}^n\frac1{i^3}+\frac1{(n+1)^3}\\ &\le\frac54-\frac1{4n}+\frac1{(n+1)^3}\\ &=\frac54+\frac{-n^3-3n^2+n-1}{4n(n+1)^3}\\ &\le\frac54+\frac{-3n^2+n-1}{4n(n+1)^3}\\ &=\frac54+\frac{-3\left(n-\frac16\right)^2-\frac{11}{12}}{4n(n+1)^3}\\

\[1+\frac 1{2^3}+\hdots+\frac 1{n^3}<1+\frac 1{2^3}+\int_2^\infty\frac 1{x^3}\mathrm dx=\frac 54\]

It is well known that $\zeta(3)<1.2021<\frac54$.