reverse taylor series

**flipperpk** · December 6th 2009, 09:31 AM

Sum the following:

$\frac{\pi^2}{2^2} - \frac{\pi^4}{2^4 3!}+\frac{\pi^6}{2^6 5!}-\frac{\pi^8}{2^8 7!}...$

We were told it had to do with the sun function. I can't figure it out. Is there a method to solving these?

**nehme007** · December 6th 2009, 10:03 AM

The given series can be written as: $\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!} (\pi/2)^{2n+2} = (\pi/2) \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!} (\pi/2)^{2n+1}$ . The sine function can be expressed as a power series as $sin(x) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$ .

If you let $x = \pi/2$ , you get that your series (with the $\pi/2$ pulled out) is just $sin(\pi/2) = 1$ , so your series converges to $\pi/2$ .

**flipperpk** · December 6th 2009, 02:13 PM

Originally Posted by nehme007

The given series can be written as: $\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!} (\pi/2)^{2n+2} = (\pi/2) \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!} (\pi/2)^{2n+1}$ . The sine function can be expressed as a power series as $sin(x) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$ .

If you let $x = \pi/2$ , you get that your series (with the $\pi/2$ pulled out) is just $sin(\pi/2) = 1$ , so your series converges to $\pi/2$ .

Can you go through the steps on solving these types of problems? I have another one that I can't figure out because it does not match up with another summation quite that nicely.

**nehme007** · December 6th 2009, 03:42 PM

You should know how to write a given series using $\Sigma$ notation and then how to manipulate the starting index of that series as needed to make the series look like the power series expansion of some popular function. Remembering the power series expansion for popular functions (trig, exponential, logs, etc.) helps so you know what you are aiming for. Also remember you can move constants into and out of a series as you please. I don't know what else to say. What's the problem you're having trouble with?

Originally Posted by flipperpk

Can you go through the steps on solving these types of problems? I have another one that I can't figure out because it does not match up with another summation quite that nicely.

**flipperpk** · December 6th 2009, 03:56 PM

$\frac{3^3}{5} - \frac{3^4}{7} + \frac{3^5}{9} - \frac{3^6}{11}...$

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