Showing that an inquality is valid

Question

Show that for every $n\ge2$, we have: $$ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{n^2}<1-\frac{1}{n} $$

My attempt:

The LHS can be simplified to:

$$\sum_{n=2}^\infty \frac{1}{n^2}.$$ The RHS term can be written as:

$$1-\sum_{n=2}^\infty \frac{1}{n},$$

On the LHS, we have that using the p-series test:

$$\sum_{n=2}^\infty \frac{1}{n^2}$$ converges, since $p>1$ for $\frac{1}{n^p}$. The RHS term $1-\sum_{n=2}^\infty \frac{1}{n}$ does not converge, since by the p-series test, the component $\frac{1}{n}$ has that $p=1$, such that this RHS is growing indefinitely.

But the RHS is actually growing towards $-\infty$, so it cannot be ever larger that the LHS.

Any ideas how I can solve this?

Thanks

The left-hand side is not $\sum_{n=2}^\infty \frac{1}{n^2}$. — Martin R, Jun 16 '23 at 08:40
Why would the RHS be written as $1-\sum_{n=2}^\infty \frac{1}{n}$? — Bruno B, Jun 16 '23 at 08:41
You can also do: $1 - \frac{1}{n} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{n-1} - \frac{1}{n} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \ldots + (\frac{1}{n-1} - \frac{1}{n}) = \displaystyle\sum_{k=2}^{k=n} \frac{1}{(k-1)k} > \displaystyle\sum_{k=2}^{k=n} \frac{1}{k^2}.$ — Adam Rubinson, Jun 16 '23 at 08:55
@AdamRubinson: That is also a solution in the suggested duplicate target: https://math.stackexchange.com/a/1496424/42969 . — Martin R, Jun 16 '23 at 08:58
@MartinR the left hand side is $\sum_{n=1}^\infty \frac1{n^2}-1$ — Luthier415Hz, Jun 16 '23 at 09:01
@Luthier415Hz: There is no infinite series on the left-hand side. It is $\sum_{k=2}^n \frac{1}{k^2}$. — Martin R, Jun 16 '23 at 09:04
@MartinR I was wrong, I thought the sum was there (invisibly). — Luthier415Hz, Jun 16 '23 at 09:05

score 3 · Accepted Answer · answered Jul 03 '23 at 16:47

I realize that the question was already posed in Proving $ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ for all $n\geq 2$ by induction, but i still wanna show my solution.

Leonhard Euler famously proved that the sum of all reciprocals of squares of natural numbers is equal to $\frac{\pi^2}{6}$, basically:
$$\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}$$ From that we can conclude that: $$\sum_{n=2}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}-1<\frac{2}{3}$$ Every term in this sum is positive. Thus we can conclude that the finite sum (starting at $n=2$) will also be smaller than $\frac{2}{3}$. This proves the inequality is true for all $n\geq3$, since $1-\frac{1}{n}\geq\frac{2}{3}$ for all $n\geq3$. The case when $n=2$ is also true, since $\frac{1}{2^2}<1-\frac{1}{2}$. Thus the inequality is true for all $n\geq2$.

Showing that an inquality is valid

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