Strong induction inequality $\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}\leq\frac{5}{8}-\frac{1}{n}$

Use strong induction to prove that

$$\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}\leq\frac{5}{8}-\frac{1}{n}$$ $$n\geq2$$

I'm not sure how to go about this. I used base cases n=2, and n=3 but I'm having trouble with the actual induction part.

I said this as my Induction Hypothesis: Suppose the claim is true for n=2, 3, ... ,k. We must show it holds for n=k+1.

Not sure where to go from there :T.

• Not that it is helpful for this question but the infinite sum $\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}+\cdots$ is $1$ less than Apéry's constant so less than $0.20205690316$ – Henry Jul 22 '18 at 17:21

Assuming it holds for $n-1$ it suffices to show that $$\frac1{n^3}\le\frac1{n-1}-\frac1n=\frac1{n(n-1)}$$ which is obviously true.
You should see that adding \begin{align*}\frac1{2^3}+\ldots\frac1{(n-1)^3}&\le\frac58-\frac1{n-1}\\ \frac1{n^3}&\le\frac1{n-1}-\frac1n\end{align*} yields $$\frac1{2^3}+\ldots\frac1{n^3}\le\frac58-\frac1n\!\!\!$$