Prove that for $ \displaystyle n≥1, \sum_ {i=1} ^{ n} \dfrac 1 {i^2} ≤ 2 - \dfrac 1 n$

Question

Can someone help me with this proof?

Prove that for $n≥1$, $\displaystyle\sum_{i=1}^n \dfrac 1 {i^2} \le 2 -\dfrac 1 n$.

The Base Case is easy but for my induction step, I get stuck at...

          (sum from i=1 to k+1 of 1/i^2) ≤ 2 - 1/(k+1)
(sum from i=1 to k of 1/i^2) + 1/(k+1)^2 ≤ 2 - 1/(k+1)

Answer 1

The induction step: Suppose the result is true when $n=k$. We show the result is true when $n=k+1$.

By the induction hypothesis, we have $$1+\frac{1}{2^2}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}\le 2-\frac{1}{k}+\frac{1}{(k+1)^2}=2-\left(\frac{1}{k}-\frac{1}{(k+1)^2}\right).$$

Now it is enough to show that $\frac{1}{k}-\frac{1}{(k+1)^2}\ge \frac{1}{k+1}$.

Equivalently, we want to show that $(k+1)^2-k\ge k(k+1)$. (We cleared denominators.) This is obvious, since $(k+1)^2-k=k^2+k+1$.

Answer 2

Note the right-hand integral estimate gives

$$\sum_{k=2}^{n}{1\over k^2}\le \int_1^n {dx\over x^2}=1-{1\over n}$$

because $f(x)={1\over x^2}$ is monotone decreasing.

Adding $1={1\over 1^2}$ gives

$$\sum_{k=1}^n{1\over k^2}$$ $$=1+\sum_{k=1}^{n-1}{1\over k^2}$$ $$\le \int_1^n {dx\over x^2}+1=1-{1\over n}+1$$ $$= 2-{1\over n}$$

Answer 3

It can also be solved by a telescoping method. \begin{align} \sum_{i=1}^n \dfrac 1 {i^2} &=1+\sum_{i=2}^n \frac{1} {i^2} \\ &\le 1+\sum_{i=2}^n \frac{1} {i(i-1)}\\ &= 1+\sum_{i=2}^{n}(\frac{1}{i-1}-\frac{1}{i} )\\ &=1+((1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots(\frac{1}{n-1}-\frac{1}{n}))\\ &=1+(1-\frac{1}{n})\\ &= 2 -\dfrac 1 n \end{align}