Show that $\frac{1}{2^2}+\frac{1}{3^2}+\text{...}+\frac{1}{n^2}<1$

Question

Show that $$\dfrac{1}{2^2}+\dfrac{1}{3^2}+\text{...}+\dfrac{1}{n^2}<1$$ for all $n\ge2,n\in N.$

Initially, we should prove the proposition is true for $n=2$. $$\dfrac{1}{2^2}\overset{?}{<}1\\\dfrac{1}{4}\overset{?}{<}1=\dfrac44$$ which is obviously true. Now let's suppose that the inequality holds for $n=k\ge2,$ $$\dfrac{1}{2^2}+\dfrac{1}{3^2}+\text{...}+\dfrac{1}{k^2}<1$$ Then it must also be true for $n=k+1$, so we have to prove $$\dfrac{1}{2^2}+\dfrac{1}{3^2}+\text{...}+\dfrac{1}{(k+1)^2}\overset{?}{<}1\\\dfrac{1}{2^2}+\dfrac{1}{3^2}+\text{...}+\dfrac{1}{(k+1)^2}=\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\text{...}+\dfrac{1}{k^2}\right)+\dfrac{1}{(k+1)^2}<1+\dfrac{1}{(k+1)^2}$$ which is actually greater than $1$.

I would like to use induction.

this is not one you can do by induction, at least not in trivial ways — Exodd, Jan 17 '22 at 17:37
Because the right side is constant and the left side is increasing, so as you saw you can't use the inequality for $n$ to prove for $n+1$. You need something stronger. — Momo, Jan 17 '22 at 17:44
@Momo, thank you! What do you mean by saying that the left side is increasing? — kormoran, Jan 17 '22 at 17:46
@Andrei - that is harder than needed. Showing it is less than $1$ is a fairly easy integration or induction. — Henry, Jan 17 '22 at 17:47

score 5 · Accepted Answer · answered Jan 17 '22 at 17:47

5

Since $j^2 > j(j-1)$ for all $j \ge 1$, then $$\frac{1}{2^2} + \cdots + \frac{1}{n^2} < \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{(n-1)n}$$ Observe $$\frac{1}{1\cdot 2} + \cdots + \frac{1}{(n-1)n} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n}\right) = 1 - \frac{1}{n} < 1$$

If you insist on giving a proof by induction, prove the stronger inequality$$\frac{1}{2^2} + \cdots + \frac{1}{n^2} <1 - \frac{1}{n}$$ for all $n \ge 2$ by induction.

answered Jan 17 '22 at 17:47

kobe

41,901

Thank you! I really don't see why $$\frac{1}{1\cdot 2} + \cdots + \frac{1}{(n-1)n} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n}\right) = 1 - \frac{1}{n} < 1$$ – kormoran Jan 17 '22 at 17:54
1

Since $\frac{1}{(k-1)k} = \frac{1}{k-1} - \frac{1}{k}$ for all $k \ge 2$, the first equality follows. The second equality follows from cancelling terms. – kobe Jan 17 '22 at 17:56
Thank you! But I still don't see how do we cancel out to get $1-\dfrac{1}{n}$. – kormoran Jan 17 '22 at 18:34
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@Medi By regrouping, the sum you’re referring to equals $$1 + \left(-\frac{1}{2} + \frac{1}{2}\right) + \left(-\frac{1}{3} + \frac{1}{3}\right) + \cdots + \left(-\frac{1}{n-1} + \frac{1}{n-1}\right) - \frac{1}{n}$$ We see that every term cancels except for the first and last term, and we obtain $1 - \frac{1}{n}$. – kobe Jan 17 '22 at 18:40

Show that $\frac{1}{2^2}+\frac{1}{3^2}+\text{...}+\frac{1}{n^2}<1$

1 Answers1