# Proof by induction of the inequality

I got problems trying to prove the following inequality by the induction method

$1+ \frac{1}{2^2}+ \ldots + \frac{1}{n^2} < \frac{7}{4}$

I've found a similar example with 2 in the right and it was recommended to prove a stronger statement, which is

$1+ \frac{1}{2^2}+ \ldots + \frac{1}{n^2} \le 2- \frac{1}{n}$

I've been trying to make up something like that, but still can't find a proper stonger statement.

Alternative approach (no induction). We show that $$\sum_{k=3}^n\frac{1}{k^2}<\frac{7}{4}-1-\frac{1}{4}=\frac{1}{2}.$$ In fact, for $n\geq 2$, $$\sum_{k=3}^n\frac{1}{k^2}\leq \sum_{k=3}^n\int_{k-1}^{k}\frac{dx}{x^2}= \int_{2}^{n}\frac{dx}{x^2}=\left[-\frac{1}{x}\right]_2^{n}=\frac{1}{2}-\frac{1}{n}<\frac{1}{2}.$$

P.S. The above inequality implies that for $n\geq 2$, $$\sum_{k=1}^n\frac{1}{k^2}\le \frac{7}{4} - \frac{1}{n}$$ which can be independently proved by induction (see cip999's answer).

• If there's a misprint in your last expression of the sum, then I understand it. I mean if k starts from 2, not 1. – GuntherOnFire Oct 18 '17 at 8:00
• @GuntherOnFire Do you mean: $\sum_{k=1}^n\frac{1}{k^2}\le \frac{7}{4} - \frac{1}{n}$ ? It is correct with $k=1$. Why do you think that we need $k=2$ here? – Robert Z Oct 18 '17 at 8:13
• Yes. If I set $n=1$, then I have $1 \le \frac{7}{4} - 1 = \frac{3}{4}$, which is incorrect. – GuntherOnFire Oct 18 '17 at 8:21
• @GuntherOnFire But I am saying that $\sum_{k=1}^n\frac{1}{k^2}\le \frac{7}{4} - \frac{1}{n}$ holds for $n\geq 2$. So $n=1$ is excluded. – Robert Z Oct 18 '17 at 9:01
• Oh, so we're just shifting the base of induction to $n=2$? I mean the steps we're taking are showing that inequality is true for the original thesis and then we're saying that we can prove a stronger inequality for the series starting from $n \ge 2$. Have I understood it correctly? – GuntherOnFire Oct 18 '17 at 9:08

$$1 + \frac{1}{2^2} + \cdots + \frac{1}{n^2} \le \frac{7}{4} - \frac{1}{n}$$ still works for $n \ge 2$.

• I think comparison with $\frac{K}{n}$ where $K \le 1$ works here, like Connor Harris did above. And this one won’t work for n equal to 1. But thanks anyway – GuntherOnFire Oct 17 '17 at 16:13
• The fact that it doesn't work for $n = 1$ isn't that much trouble, since it can be easily checked by hand that the original thesis is true even in that case. Then, of course, you should be taking $n = 2$ as the base case of the induction. – cip999 Oct 17 '17 at 17:38
• Oh, I see. So we check that for n=1 our original thesis is true and then we prove that the stronger inequality make sense for all $n \ge 2$. – GuntherOnFire Oct 18 '17 at 7:57