Sum of Series Homework: Find \sum^{\infty}_{0}\frac{(-9)^{n}}{(2n+1)!}

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Homework Statement


Find the sum of the series \sum^{\infty}_{0}\frac{(-9)^{n}}{(2n+1)!}


Homework Equations


Alternating series Estimation Theroem


The Attempt at a Solution


I think it satisfies the conditions necessary for an alternating series to converge. The limit as n approaches infinity of b_{n}=0, and b_{n}>b_{n+1}. So listing out the terms I get 1-(9/6)+(81/120)-(720/5040)+... But at this point I'm not sure how far to expand the series, because the problem just says find the sum of the series, which makes me think that there would be an exact answer, but I can't think of another way to find a sum. It's not a geometric sequence, and I don't think I could estimate it with integration either.
 
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Not sure how you are supposed to find the answer. If you replace the -9 with (-1)^n * (9)^n it looks close to the series expansion for sine. But you want n^2 +1 in the numerator power. Change the 9 to 3^2 gives you (3)^2n. Multiply outside the sum by 1/3 and inside by 3 gives you (3)^(2n+1). So you get (1/3) * sin(3).
 
Does this

\sum^{\infty}_{n=0}<br /> \frac{(-1)^{n} \cdot x^{2n+1} }{(2n+1)!}<br />

remind you of, oh, say, any particular Maclaurin series you may have seen before...? (It converges for any real x. Many problems in infinite series are set where you're expected to recognize what function the series represents in order to evaluate it...)

Now, what you have is not quite the same, but that can be easily fixed. Write out the first few terms of this series and of your series. In comparing the two, you should notice what must be done to express your series in terms of a multiple of a familiar function...


Chrisas said:
Not sure how you are supposed to find the answer. ...

What you described is pretty much what the student is asked to notice. But, here at PF, you really shouldn't just hand out the answer as well...
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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