How do you use a power series to find the exact value of the sum of the series #1-pi^2/(2!)+pi^4/(4!) +pi^6/(6!) + …# ?

Answer 1

Charles Anctil

Since

#1-x^2/{2!}+{x^4}/{4!}-x^6/{6!}+cdots=cosx#,

#1-pi^2/{2!}+pi^4/{4!}-pi^6/{6!}+cdots=cos(pi)=-1#

I know that it is different from what you have, but I have a feeling that this is what you meant to ask.

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Answer 2

Luke Alvarez

To find the sum of the series (1 - \frac{\pi^2}{2!} + \frac{\pi^4}{4!} + \frac{\pi^6}{6!} + \ldots), you can use the power series expansion of the function (e^x). Specifically, you can use the Taylor series expansion of (e^x) centered at (x = 0), which is:

[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}]

If you substitute (x = \pi) into this series, you get:

[e^\pi = \sum_{n=0}^{\infty} \frac{\pi^n}{n!}]

Now, observe that the series you're given is a rearrangement of this series. So, the sum of the given series is simply (e^\pi).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.