Difficult induction problem

Question

Been trying to figure this one out for a couple of days to no avail. Appreciate if someone could set me in the right direction.

Prove by induction that $$2\left(1+\frac18+\frac1{27}+...+\frac1{n^3}\right)<3-\frac1{n^2}\text{ for all }n\geq 2.$$

Answer 1

I guess the base case is cleared.

Assume the statement is true for some $n\in\mathbf{N}$. Then $$ \begin{aligned} 2\Bigl(1+\frac{1}{8}+\cdots+\frac{1}{n^3}+\frac{1}{(n+1)^3}\Bigr)&<3-\frac{1}{n^2}+\frac{2}{(n+1)^3}.\\ &=3-\frac{(n+1)^3-2n^2}{(n+1)^3n^2}\\ &<3-\frac{3n^2-2n^2}{(n+1)^3n^2}\\ &=3-\frac{1}{(n+1)^3} \end{aligned} $$ In the second inequality, I used $(n+1)^3=n^3+3n^2+3n+1>3n^2$.

Answer 2

Let the LHS be denoted by $f(n)$ and the RHS by $g(n)$. Then $$ f(n+1)-f(n)=\frac{2}{(n+1)^3}<g(n+1)-g(n)=\frac{2n+1}{n^2(n+1)^2}=\frac{2+\frac 1n}{n(n+1)^2} $$ which should provide you the inductive step.

Answer 3

$$ 0 \lt 3n+1 \\ 2n^2 \lt 2n^2+3n+1 = (2n+1)(n+1) $$ so $$ \frac{2n^2}{n^2(n+1)^3} \lt \frac{(2n+1)(n+1)}{n^2(n+1)^3} $$ or $$ \frac{2}{(n+1)^3} = \lt \frac{2n+1}{(n+1)^2n^2} = \frac1{n^2} - \frac1{(n+1)^2} $$ hence $$ 2\sum_{k=1}^{n+1}\frac1{k^3} = 2\sum_{k=1}^{n}\frac1{k^3} +\frac2{(n+1)^3} \lt 3-\frac1{n^2} + (\frac1{n^2} - \frac1{(n+1)^2}) \\ = 3 - \frac1{(n+1)^2} $$

Answer 4

Base Case is easy. Try for 2(1 + 1/8 + 1/27 ... 1/n^3 + 1/((n+1)^3)) < 3 - 1/(k^2)

Let original LHS equal a

then new LHS is a + 2/((k+1)^3) , which is < 3 - 1/(k+1)^2

multiply everything by (k+1)^3, and subtract the 2 from the right

a(k+1)^3 + 2 < 3(k+1)^3 - k - 1

a(k+1)^3 < 3(k+1)^3 - k - 3

Divide again by (k+1)^3 (probably not that efficient but it's what made sense to me while solving)

a < 3 - (k+3)/(k+1)^3

That RHS must logically be greater than 3 - 1/(k)^2 for k>2 because then since a is known to be less than 3 - 1/(k^2), due to transitive property, a will be less than that as well and the inequality above this paragraph will be proven true

this proving is easily done

3- (k+3)/(k+1)^3 <? 3 - 1/(k)^2

simplify multiply by -1 and flip signs

(k+3)/(k+1)^3 >? 1/(k)^2

for k = 2, RHS is greater, so 3 - RHS < 3 - LHS, so by PMI the argument is true