Upper estimate for partial sums of the series $\sum 1/n^3$

Question

I'm looking for a proof of: $$1 + {1\over 8} + {1 \over 27} + \dots + {1 \over n^3} < 1.5 − {1 \over n}$$ for all integers $n>2$.

I have been working on proof by mathematical induction for a few weeks now. It is tricky but I can usually get the correct result but I encountered this problem using an inequality and of course I can test the base case but I can't figure out for the life of me do the rest.

Answer 1

So, we have established the base case, which is $$1 + \frac18 + \frac{1}{27} < 1.5-\frac13$$ Now we assume we know that the inequality holds for all $n$ with $2<n \leq k$, for some $k\geq 3$, and we need to examine what happens for $n = k+1$, which means to examine this sum: $$ 1+\frac18+\frac{1}{27} +\cdots + \frac1{k^3} + \frac{1}{(k+1)^3} $$ Now, we have assumed for a fact (the induction hypothesis) that the sum of all terms except the last one is less than $1.5-1/k$: $$ \color{blue}{1+\frac18+\frac{1}{27} +\cdots + \frac1{k^3}} + \frac{1}{(k+1)^3} < \color{blue}{\frac32 - \frac{1}{k}} + \frac{1}{(k+1)^3} $$ and we want to compare this to $1.5-1/(k+1)$, i.e. $$ \frac32 - \frac{1}{k} + \frac{1}{(k+1)^3} \overset?<\frac32-\frac{1}{k+1}\\ \frac{1}{k} -\frac{1}{(k+1)^3} \overset?>\frac{1}{k+1} $$ and since we have assumed $k \geq 3$, this is true.

Answer 2

Hint: show that is true for $n=3$. Suposse that is true for $n=k$, and then proof that is true for $n=k+1$, $$1 + {1\over 8} + {1 \over 27} + \dots + {1 \over k^3} + {1 \over (k+1)^3}< 1.5 − {1 \over (k+1)}$$ is same $$1 + {1\over 8} + {1 \over 27} + \dots + {1 \over k^3} + {1 \over (k+1)^3}< 1.5− {1 \over k}+ {1 \over k}− {1 \over (k+1)}$$ Then proof that $${1 \over (k+1)^3}< {1 \over k}− {1 \over (k+1)}$$

Answer 3

Interesting question.
The limit as $n\to \infty$ is $\zeta(3)$ or Apery's constant.
The inequality to be proven shows that an upper bound for this is $1.5$.
(Apery's constant is actually $\zeta(3)=1.2020569031\cdots $)

The derivation below uses a direct approach instead of induction, and is provided for reference only.

First we note that

$$\begin{align} \color{blue}{\frac 1{2^3}+\int_2^n\frac 1{x^3}dx}-\color{green}{\int_2^n\frac 1{x^2}dx} &=\frac 18+\left(\frac 18-\frac 1{2n^2}\right)-\left(\frac 12-\frac 1n\right)\\ &=-\frac 14-\frac 1{2n^2}+\frac 1n\\ &=-\frac {n^2-4n+2}{4n^2}\\ &=-\frac {(n-\alpha)(n-\beta)}{4n^2}\qquad&&\small{\text{where $\alpha=2-\sqrt 2,\beta=2+\sqrt2$}}\\ &<0\qquad &&\text{for}\quad n>3 \;(n\in\mathbb N)\\ \Longrightarrow \color{blue}{\frac 1{2^3}+\int_2^n\frac 1{x^3}dx}&<\color{green}{\int_2^n\frac 1{x^2}dx}\\ \end{align}$$ Also, $$\begin{align} \color{orange}{\sum_{r=2}^n\frac 1{r^3}}<\color{blue}{\frac 1{2^3}+\int_2^n\frac 1{x^2}dx}\end{align}$$ Hence $$\begin{align} \color{orange}{\sum_{r=2}^n\frac 1{r^3}}&<\color{blue}{\frac 1{2^3}+\int_2^n\frac 1{x^3}dx}<&&\color{green}{\int_2^n\frac 1{x^2}dx}\\ \sum_{r=2}^n\frac 1{r^3}&<&&\int_2^n\frac 1{x^2}dx\\ 1+\sum_{r=2}^n\frac 1{r^3}&<&&1+\int_2^n\frac 1{x^2}dx\\ \sum_{r=1}^n\frac 1{r^3}&<&&1+\left[-\frac 1x\right]_2^n=1+\frac 12-\frac 1n\\ \sum_{r=1}^n\frac 1{r^3}&<&&1.5-\frac 1n\qquad\blacksquare \end{align}$$ It can be shown by hand that this inequality holds for $n=3$.
Hence the inequality holds for $n>2 (n\in\mathbb N)$