Proofs for 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots=\frac{\pi^2}{6}

[murshid_islam]

can anyone help me with the proofs:

$$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots=\frac{\pi^2}{6}$$

if $$F_i$$ is the ith Fibonacci number, then

$$F_1+F_2+F_3+\ldots+F_n=F_{n+2}-1$$

$$F_2+F_4+F_6+\ldots+F_{2n}=F_{2n+1}-1$$

$$F_1+F_3+F_5+\ldots+F_{2n-1}=F_{2n}$$

$$F_1^2+F_2^2+F_3^2+\ldots+F_n^2=F_nF_{n+1}$$

I proved the last four using induction. But how can i prove them without using induction?

 

[lurflurf]

can anyone help me with the proofs:

$$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots=\frac{\pi^2}{6}$$

if $$F_i$$ is the ith Fibonacci number, then

$$F_1+F_2+F_3+\ldots+F_n=F_{n+2}-1$$

$$F_2+F_4+F_6+\ldots+F_{2n}=F_{2n+1}-1$$

$$F_1+F_3+F_5+\ldots+F_{2n-1}=F_{2n}$$

$$F_1^2+F_2^2+F_3^2+\ldots+F_n^2=F_nF_{n+1}$$

I proved the last four using induction. But how can i prove them without using induction?

euler found your sum zeta(2) by writing
sin(x)=x(1-x^2/(k pi)^2)(1-x^4/(k pi)^4)(1-x^6/(k pi)^6)...
and expanding the product and setting it equal x-x^3/3!+x^5/5!+...
It is often done as a routine exercise with Fourier series.
There are many ways to prove it.
see
http://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html
for the other 4 what's wrong with induction?
Fn=(a^n-b^n)/(a-b)
where a=.5(1+sqrt(5)) b=.5(1-sqrt(5))
so you could express those sums as geometric series among other methods
see
http://mathworld.wolfram.com/FibonacciNumber.html

 

[murshid_islam]

for the other 4 what's wrong with induction?

there's nothing wrong with induction. but i wanted get the right hand sides from the left hand sides (from scratch), if you know what i mean.

anyway, thanks very much for the help. and additional information is always welcome.