How do I run an infinite series by creating a function?

sum(1/n^2), n=1:infinite

Answer 1

[Ameer Hamza]

Only possible if you have symbolic toolbox

syms n
y = symsum(1/n^2, 1, inf);

Result

>> y
y =
pi^2/6

Otherwise you will need to numerical approximation using a finite number of elements

n = 1:10000;
y = sum(1./n.^2);

Result

>> y
y =
    1.6448

[Khalid Ben Houria]

thanks my friend and this serie : 1+(1/3^2)+(1/5^2)+...........+(1/n^2)+........ how?

by function?

function Y=serieinf(n) or what?

[Ameer Hamza]

You can define the denominator like this

syms n
y = symsum(1/(2*n-1)^2, 1, inf);

[Cory Parker]

Basel Problem