Help with proof using induction: $1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ [duplicate]

I am having trouble with the following proof:

For every positive integer $n$: $$1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$$

My work: I have tried to add $\frac{1}{(k+1)^2}$ to $2-\frac{1}{k}$ in the inductive step and reduce it down to $2-\frac{1}{k+1}$ but cannot do so. I am beginning to think that my entire approach is wrong.

Solution 1:

Hint: $\frac{1}{n^2} < \frac{1}{(n-1)} - \frac1n$, and telescope.

Solution 2:

Inductive Hypothesis: suppose $\sum_{k=1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$.

Inductive Step: then suppose $\sum_{k=1}^{n+1} \frac{1}{k^2} \leq 2 - \frac{1}{n} + \frac{1}{(1+n)^2}. $

So it suffices to show that $ - \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$.

But this, by simple algebra, is equivalent to $ n \leq n+1 $, which is obviously true $\forall n \in \mathbb{N}$.

EDIT Proof that $ - \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$ is equivalent to $n \leq n+1$.

Add $\frac{1}{n}$ to both sides to se that what we need to show is $\frac{1}{(1+n)^2} \leq \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}.$ Multiply both sides by $(n+1)$. Then take reciprocals and remember to swap the inequality sign.

Solution 3:

In the induction step observe that $1$+$1/4$ +...+$1/k^{2}$ + $1/(k+1)^{2}$$\leq$ $2$- $1/k$ + $ 1/(k+1)^{2}$

Now $1/(k+1)^{2}$ $\leq$ $1/k(k+1)$=$1/k$-$1/(k+1)$. Use this inequality in previous to obtain the answer.

Solution 4:

Others have commented on using induction. Here's another approach just 'cause:

We know the infinite sum is equal to $\frac{\pi^2}{6}$ which is approximately $1.644$, so we know it is true for $n > 2$ (infinite sum > partial sum with all positive terms)

So we can just hand check $n = 1, 2$: Indeed, $1 \leq 1$ and $\frac{5}{4} \leq \frac{3}{2}$