1 + 1/4 + 1/9 + 1/76 + ......... + 1/n² 2, n E N

The Question Should be like
1+1/4+1/9+1/16+................+ 1/n²<2-1/n    for n>=2


Proof:
Since n = 2 being the base case,for n=2
1+1/4=5/4=1.25
and for n=2
2-1/2=3/2=1.5
clearly, 1.25 < 1.5
So the given formula is true for n=2

Let the formula is true for n,
So, we have
1 + 1/4 + 1/9 + 1/6 + ... + 1/n² < 2 - 1/n.................(1)

Now,we have to show that it is true for n+1, i.e.
1 + 1/4 + 1/9 + 1/6 + ... + 1/n² + 1/(n + 1)² < 2 - 1/(n + 1).
First we take
1 + 1/4 + 1/9 + 1/6 + ... + 1/n² + 1/(n + 1)²
= (1 + 1/4 + 1/9 + 1/6 + ... + 1/n²) + 1/(n + 1)²
< 2 - 1/n + 1/(n + 1)² ( putting value from (1))
= 2+$\frac{-(n+1{)}^2+n}{n(n+1{)}^2}$
=$2+\frac{-(n^2+2n+1)+n}{n(n+1{)}^2}$
=$2-\frac{n^2+n+1}{n(n+1{)}^2}$
From here, we just need to show that:
2 - (n² + n + 1)/[n(n + 1)²] < 2 - 1/(n + 1) for all n ≥ 2.

Solving this inequality:
2 - (n² + n + 1)/[n(n + 1)²] < 2 - 1/(n + 1)
=> -(n² + n + 1)/[n(n + 1)²] < -1/(n + 1)     (Cancelling 2 from both sides)
=> (n² + n + 1)/[n(n + 1)²] > 1/(n + 1) (Cancelling minus sign,then < is changed to >,according to rule)
=> (n² + n + 1)/[n(n + 1)²] - 1/(n + 1) > 0
=> [(n² + n + 1) - (n² + n)]/[n(n + 1)²] > 0  (Taking LCM and solvin)
=> 1/[n(n + 1)²] > 0.

Since this holds for every n > 0.
So,We have shown that:
2 - (n² + n + 1)/[n(n + 1)²] < 2 - 1/(n + 1) for all n ≥ 2.

Thus, (1 + 1/4 + 1/9 + 1/6 + ... + 1/n²) + 1/(n + 1)²
< 2 - 1/n + 1/(n + 1)²
= 2 - (n² + n + 1)/[n(n + 1)²]
< 2 - 1/(n + 1).

So, the formula is true for n+1, hence by principal of matehematical induction, it will be true for all values n>=2