# Sum the Taylor Series

1. Dec 5, 2009

### vucollegeguy

The series is:

(33/5) - (34/7) + (35/9) - (36/11)+...

Looking at this, I'm guessing I can use the Taylor Series for arctan(x) but I dont know how to apply it or where to begin.

Any help is greatly appreciated.

2. Dec 5, 2009

### Staff: Mentor

You have 33/5 - 34/7 + 35/9 - 36/11 +...
= 33(1/5 - 3/7 + 32/9 - 33/11 +...)
$$=~3^3~\sum_{n = 0}^{\infty}(-1)^n~\frac{3^n}{2n + 5}$$

The series above is an alternating series. Do you know a test for determining whether such a series converges?

3. Dec 5, 2009

### vucollegeguy

I can use the alternating series test where I let bn=(3n)/(2n+5), right?

4. Dec 5, 2009

Right

5. Dec 5, 2009

### LCKurtz

The title of your post is "sum the series". I doubt that is what you mean. You probably mean test it for convergence.

Anyway, remember the alternating series test will only tell you a series is convergent. If the test for convergence fails, that does not tell you the series diverges. So the alternating series might not (hint, hint) be the end of the story for this problem.

Last edited: Dec 5, 2009
6. Dec 6, 2009

### vucollegeguy

This is what I'm confused about. This is a practice exam for my final. And question specifically says "Sum the following series."

I did what mark44 did and factored out the 3^(3) but I didn't come up with the sum formula. When I factored it out, it looked as if it were from the Taylor Series of arctan(x).

Now, I don't know where to go from here.

7. Dec 6, 2009

### Count Iblis

If you forget about convergence issues and want to "sum" the series in the sense of taking the Taylor expansion of some function and insert some value for x so that the expansion matches your series, then this is an example of a "resummation" method.

The idea is then that the series represents a finite number that was derived formally correctly, but it is an expansion around some point yielding a divergent series. But the terms of the expansion will contain all the information about the number which you have to "decode".

You are on the right track with the arctan function. If you look at two successive terms and forget about the missing terms at the start, what should you take for x?