What does it mean to "sum a series"?

**Mattpd** · April 29th 2010, 04:41 PM

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

**skeeter** · April 29th 2010, 05:02 PM

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 ?

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

**Mattpd** · April 29th 2010, 05:30 PM

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)

**skeeter** · April 29th 2010, 05:33 PM

Originally Posted by Mattpd

it looks to be cos(x)

correct ... so, what is the sum of that series?

**Mattpd** · April 29th 2010, 05:38 PM

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?

**skeeter** · April 29th 2010, 05:45 PM

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?

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

**Mattpd** · April 29th 2010, 05:50 PM

Pi/3?

**skeeter** · April 29th 2010, 06:06 PM

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

**Mattpd** · April 29th 2010, 06:30 PM

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?

**HallsofIvy** · April 30th 2010, 03:56 AM

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".

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