Finding the sum of a series in calc

[karadda]

Homework Statement

http://imgur.com/KE9lZ.png"

I'm reviewing for my calculus II exam, and I'm unable to solve any problems of this type "find the sum of the series." sometimes i'll be given a series in sigma notation (like this problem), and sometimes i'll be given the first few terms of the series. Either way, I'm completely lost here. I could use a step by step guide on how to solve this type of problem. Thanks!

 

[gabbagabbahey]

The general method is to compare the series to certain well-known series, like geometric series, binomial series, or Taylor series of exponentials or trig functions.

For your specific example, start by re-writing it as

$$\sum_{n=0}^{\infty}\frac{(-1)^n\pi^{2n}}{6^{2n+1}(2n)!}=\frac{1}{6}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\left(\frac{\pi}{6}\right)^{2n}$$

Can you think of a Taylor series of a simple function, that only contains even powers and alternating terms?

 

[hgfalling]

My first step would be to try to identify what kind of series this is, based on the types of series that you know and have studied in your class.

Presumably you are familiar with the geometric series:
$$\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}, |x| < 1$$

Geometric series are easy to identify and easy to evaluate.

If that doesn't work, try to identify the series as the Taylor series of some function or another.

It's probably useful to make sure you have a clear understanding of a) how to expand a function in a Taylor series and b) know the Taylor series for some common functions, such as ln (1+x), $e^x$, $\sin x$, $\cos x$, etc.

Usually if there's a factorial in the denominator, it's a Taylor series.

There are also some other notable series results, such as:

$$\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$$ and the like.

Among these three categories, I would imagine that you'll find almost all the series that you would be asked on a second-semester calculus exam. Maybe I've missed something and other commentors will point it out.

In any event, the series you ask about is covered somewhere in the above. :)

 

[karadda]

Can you think of a Taylor series of a simple function, that only contains even powers and alternating terms?

f(x) = cos x

Though its not exactly what I'm looking for, it matches that criteria... I think. I still don't quite understand why these comparisons with well known series is useful. I find a series like the one I have, manipulate it a bit to look like it..then integrate an improper integral?

 

[gabbagabbahey]

f(x) = cos x

Though its not exactly what I'm looking for, it matches that criteria... I think. I still don't quite understand why these comparisons with well known series is useful. I find a series like the one I have, manipulate it a bit to look like it..then integrate an improper integral?

Well, what is the Taylor series for $\cos(x)$?

 

[karadda]

$$\Sigma$$ (-1)^n x^(2n) / (2n)!
from n = 0 to infinity

It's almost exactly like my integral except for the 1/6 out front and the pi/6 inside.

does that mean pi/6 is the x value?

1/6 cos (pi/6) = $$\sqrt{3}$$ / 12

 

[gabbagabbahey]

Yes, since $$\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$$, you know that

$$\cos\left(\frac{\pi}{6}\right)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\left(\frac{\pi}{6}\right)^{2n}$$

And so your series is just $\frac{1}{6}\cos(\pi/6)=\frac{\sqrt{3}}{12}$

 

[karadda]

! thank you both very much. I'm going to go practice a few (lots) more of these.

 

[Mimmik617]

Thanks guys. I have been looking all over trying to find how to do this same thing. I'm taking Calculus 2 online right now for my senior year, and this subject has been very vague. Appreciate it!