# Euler's equalities

1. Nov 29, 2009

### coki2000

Hello,
Can you explain to me why

$$(1-1+1-1...)=\sum_{n=0}^{\infty}(-1)^n=\frac{1}{2}$$

and

$$(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}...)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi ^2}{6}$$

I don't understand these equalities.Thanks.

2. Nov 29, 2009

### HallsofIvy

The first is being treated as a geometric series:
$$\sum_{n=0}^1 r^n= \frac{1}{1-r}$$
Taking r= -1 gives 1/(1-(-1))= 1/2.

Of course, that formula is really only valid for |r|< 1. And by the normal definition of "sum of an infinite series", limit of partial sums, the sequence of partial sums is 1, 0, 1, 0, ... which doesn't converge. So that's strectching a point, at least.

3. Nov 29, 2009

### coki2000

But why do we use this formula because this is wrong for r=-1.And why do you take the average of the partial sums(1,0,1...).Thanks for your helps.

4. Nov 29, 2009

### HallsofIvy

Because Euler decided to do it! As I said, that is NOT a valid calculation using normal definitions. Euler was famous for "playing fast and loose" with sequences and series- and getting useful, if not "correct" results.

5. Nov 30, 2009

### coki2000

Okey.Thanks.
What is the solution of second equality?

$$(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}...)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi ^2}{6}$$

6. Nov 30, 2009