Thread: What does it mean to "sum a series"?

1. What does it mean to "sum a series"?

The series is: $\sum (-\pi ^2)^n/(2n)!9^n$

and the instruction are to "sum the series." What is expected of me here? I am supposed to make an estimation? Treat it as an alternating series? I'm not sure where to start

2. Originally Posted by Mattpd
The series is: $\sum (-\pi ^2)^n/(2n)!9^n$

and the instruction are to "sum the series." What is expected of me here? I am supposed to make an estimation? Treat it as an alternating series? I'm not sure where to start
what function does this series represent ?

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

3. Originally Posted by skeeter
what function does this series represent ?

$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$
it looks to be cos(x)

4. Originally Posted by Mattpd
it looks to be cos(x)
correct ... so, what is the sum of that series?

5. Originally Posted by skeeter
correct ... so, what is the sum of that series?
Not sure. I take it I should have it memorized? Or solve it by approximation like other alternating series?

6. Originally Posted by Mattpd
Not sure. I take it I should have it memorized? Or solve it by approximation like other alternating series?
you just said that $\sum_{n = 0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = \cos{x}$ , right?

LOOK at the your series again ... what is "x" in that series?

$\sum_{n = 0}^{\infty} \frac{(-1)^n \textcolor{red}{\pi}^{2n}}{(2n)! \cdot \textcolor{red}{3}^{2n}}$

7. Pi/3?

8. Originally Posted by Mattpd
Pi/3?
this was strictly a "recognition" problem for an infinite series representation of $\cos{x}$ at a specific value of $x$ ...

$\cos\left(\frac{\pi}{3}\right)$ = the sum of the series

9. Thank you millions! So the sum of the cos function is x, and this function is the same as the cos function except with a different x value?

10. Originally Posted by Mattpd
Thank you millions! So the sum of the cos function is x, and this function is the same as the cos function except with a different x value?
This makes no sense at all! "cos function" is not a sum and so there is no "sum of a cos function". And x is the argument of the function, not the "sum".