What is the result of this sum: 1+2+3+4+5+...=? I know only experts get the right answer. I am looking for them

Here we use Riemann Zeta function:

Zeta(s)= 1/1^s + 1/2^s + 1/3^s + ...

Taking s= -1 we get:

Zeta(-1) = 1 + 2 + 3 + 4+ ... and Zeta(-1) = -1 / 12

What is wrong here?

Answer is: -1/12 using Riemann Zeta function

1+2+3+4+5+.....n=n(n+1)/2=(n²+n)/2

but 1+2+3+4+5.......=+infinity

1+2+3+4+...+n=(1+2+3+...+n+n+n-1+....+2+1)/2 = n(n+1)/2

best answer for this question is 

INFINITY

infinity.because it is continued

am I right?

sum of N natural numbers is  N * (N + 1) / 2 . Just put your N value in the formula and calculate the answer. Thanks.

Riemann Zeta is working only for complex numbers where real part of "s" is greater than 1. "s" can't be "-1". you are trying to mix two unmixables theories. Patrick Kintu has given you correct answer.

n(n+1)/2

1(1+1)/2 = 1 , 2(2 + 1)/2

one upon Twelve in Negative

n(n+1)/2 n=last digit as the digits are in AP