How to immediately see that $2-\frac{(n+1)^2-n}{n(n+1)^2}\le2-\frac1{n+1}$

Question

My question concerns the following proof:

Prove $\sum_{k=1}^n 1/k^2\le2-1/n$ for every $n$.

Proof. Clearly this holds for $n=1$. Assume it holds for some $n\ge1$. Then $\sum_{k=1}^{n+1}1/k^2\le2-1/n+1/(n+1)^2=2-\frac{(n+1)^2-n}{n(n+1)^2}\le2-1/(n+1)$. The proof is complete.

So, my question is how does one immediately see that $$2-\frac{(n+1)^2-n}{n(n+1)^2}\le2-\frac1{n+1}\ ?$$ Is there a quick logical way to surmise that the expression on the left must be less than or equal to the expression on the right or is there a bit more exploration required to see that this is true? (i.e. plugging in a few values, etc.)

Answer 1

This is equivalent to $$\frac{(n+1)^2-n}{n(n+1)^2} \ge \frac{1}{n+1}$$ Since $n+1>0$, we can cancel it: $$\frac{(n+1)^2-n}{n(n+1)} \ge 1$$ Since $n(n+1)>0$, this is equivalent to $$(n+1)^2-n \ge n(n+1)$$ which you can see to be true by expanding both sides.

Answer 2

$$S_n=\sum_{k=1}^{n}\frac{1}{k^2}=1+\sum_{k=2}^{n}\frac{1}{k^2}\leq 1+\sum_{k=2}^{n}\frac{1}{(k-1)k}$$ and since $\frac{1}{(k-1)k}=\frac{1}{k-1}-\frac{1}{k}$, the last sum is a telescopic sum and $$ S_n \leq 1+\left(1-\frac{1}{n}\right)=2-\frac{1}{n} $$ as wanted.

Answer 3

$$2-\frac{(n+1)^2-n}{n(n+1)^2}\le2-\frac1{n+1}\ \implies\frac{(n+1)^2-n}{n(n+1)^2}\ge\frac{1}{n+1}$$ Rewriting the equation: $$\frac{\frac{(n+1)^2}{n(n+1)}-\frac{n}{n(n+1)}}{n+1}\ge\frac{1}{n+1}$$ $$\frac{1+\frac{1}{n}-\frac{1}{n+1}}{n+1}\ge\frac{1}{n+1}\implies1+\frac{1}{n}-\frac{1}{n+1}\ge1$$This is true for $n\ge1$