Please help!! infinite sequences and series...

the equation: summation sign (upper limit:infinity, lower limit: n=1) 1/n^2=1/1 + 1/4 + 1/9 + 1/16 + 1/25...=pi^2/6 was established by euler.

a) use this formula to evaluate the sum of the even terms, summation sign (upper limit:infinity, lower limit: n=1) 1/(2n)^2= 1/4+ 1/16+ 1/36+ 1/64+ 1/100+...

b) now evaluate the sum of the odd terms

Hello,

The following relationship is a given:
∑(n=1→+∞) 1/n² = π²/6

Then:
∑(n=1→+∞) 1/(2n)² = ∑(n=1→+∞) 1/4n² = ¼ ×∑(n=1→+∞) 1/n² = ¼ × π²/6 = π²/24

And also:
∑(n=0→+∞) 1/(2n+1)² = [∑(n=1→+∞) 1/n²] - [∑(n=1→+∞) 1/(2n)²] = π²/6 - π²/24 = π²/8

Logically,
Dragon.Jade :-)

as per Euler :
∞
Σ 1 / n ² = π ² / 6
n=1


so sum of the even numbers is:
∞ .................. ∞
Σ 1 / ( 2n ) ² = Σ 1 / 4 n ²
n=1 ............... n=1

...... ∞
= ¼ Σ 1 / n ² = ¼ ( π ² / 6 )
..... n=1

= π ² / 24
░░░░░░░░


sum of the odd numbers = Euler's sum – sum of even numbers;
..... = ( π ² / 6 ) – ( π ² / 24 )
..... = ( 4 π ² / 24 ) – ( π ² / 24 )
..... = 3 π ² / 24
..... = π ² / 8
░░░░░░░░░

(a)The sum of the even terms is 1/2^2 + 1/4^2 + 1/6^2+....
= (1/2^2)[1/1+1/2^2+1/3^2+....)=(1/2^2)(pi…

(b) Sum of odds =pi^2/6 - pi^2/24=3pi^2/24=pi^2/8