What If It Isn't Infinite?

First, let

$s_1 = 1-1+1-1+...$

then

$1-s_1 = 1 - (1+1-1+1-1+...) = s_1$

so that

$s_1 = \dfrac{1}{2}$

Now, let

$s_2 = 1+1+1+1+...$
$2s_s = 0+2+0+2+...$

Subtracting one from the other, we have

$s_2 - 2s_2 = 1-1+1-1+...=s_1= \dfrac{1}{2}$

so that

$s_2 = -\dfrac{1}{2}$

Now, let

$s_3 = 1+2+3+4+...$
$4s_3 =0+4+0+8+...$

Subtracting one from the other, we have

$s_3-4s_3 = 1-2+3-4+...$

But since

$\dfrac{1}{(1+x)^2} = 1-2x+3x^2-4x^3+...$

Then for $x=1$ , we have

$\dfrac{1}{(1+1)^2} =\dfrac{1}{4} = 1-2+3-4+...$

so that

$s_3-4s_3 = \dfrac{1}{4}$

or

$s_3 = -\dfrac{1}{12}$

Thus

$-s_2 + 6s_3 = 5+11+17+23+... = 0$

As another way of checking this,

$-Zeta(0)+6Zeta(-1) =0$

where

$Zeta(s) = \displaystyle\sum _{ p=1 }^{ \infty }{ \frac { 1 }{ { p }^{ s } } } = \dfrac{1}{1^s} + \dfrac{1}{2^s} + \dfrac{1}{3^s} + \dfrac{1}{4^s} + ...$

Still another way of checking this is

$5+11+17+23+... =\displaystyle\sum _{ k=1 }^{ n }{ \left( 6k-1 \right) } = 2n+3n^2$

Integrating this from $n=-1$ to $0$ , we have

$\displaystyle\int _{ -1 }^{ 0 }{ \left( 2n+3{ n }^{ 2 } \right) dn } =0$

Check Mathologer for the correct approach to this.

Also check out Sums of Divergent Series

Given series is : $6n-1$

$\ce {s3}=sum_{n=1}^\infty 6n-1= 6(1+2+3+\ldots)-(1+1+1+\ldots)$

Michael Mendrin demonstrates in above solution:

$\ce {s1} =1+2+3+\ldots=-\frac {1}{12}$

$\ce {s2} =1+1+1+\ldots=-\frac {1}{2}$

$\therefore \ce{s3}= 6 \ce{s1} - \ce{s2} =6 \times -\frac {1}{12} +\frac {1}{2} =0$