Induction of $\sum^n_{k=2} 1 - 1/k^2 =(1+n)/(2n)$

Question

Can someone help me with this induction please?

$$\sum^n_{k=2} 1 - 1/k^2 ={1+n\over 2n}.$$

Answer 1

Hint

$$\sum^{n+1}_{k=2} 1 -\frac{1}{k^2}=\left(\sum^{n}_{k=2} 1 -\frac{1}{k^2} \right) +1-\frac{1}{(n+1)^2}$$

Once you fix your problem (note that the sequence on the LHS is increasing approximately ar the rate of $S_n \sim n$, while the RHS is tending to a constant, you probably copied the problem wrong) just prove $P(2)$ and use the above formula to prove the inductive step.

P.S. Are you sure the exercise is NOT

$$\prod^{n}_{k=2}( 1 -\frac{1}{k^2})=\frac{n+1}{2n}$$

if it is, note $$\prod^{n+1}_{k=2} (1 -\frac{1}{k^2})=(1 -\frac{1}{(n+1)^2})\prod^{n}_{k=2} (1 -\frac{1}{k^2})=(\frac{n(n+2)}{(n+1)^2})\prod^{n}_{k=2} (1 -\frac{1}{k^2}) $$

Answer 2

Assuming that this is a summation and not a product, if you enjoy generalized harmonic numbers, $$S_n=\sum^n_{k=2}\left( 1 - \frac 1 {k^2} \right)=n-H_n^{(2)}$$ For large values of $n$, this would give $$S_n=-\frac{\pi ^2}{6}+n+\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ which has little to do with $\frac{n+1}{2n}$.

It must be the product as already said in comments and answers.