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

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SUMMARY

The series \(\sum^{\infty}_{0}\frac{(-9)^{n}}{(2n+1)!}\) converges as it meets the criteria for an alternating series, specifically that the limit of \(b_{n}\) approaches zero and \(b_{n} > b_{n+1}\). The series can be transformed to resemble the Maclaurin series for sine, leading to the conclusion that the sum can be expressed as \(\frac{1}{3} \sin(3)\). This transformation involves recognizing the series' structure and adjusting the terms accordingly to relate it to known functions.

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Homework Statement


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


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[tex]_{n}[/tex]=0, and b[tex]_{n}[/tex]>b[tex]_{n+1}[/tex]. 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

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

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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