Questions: Compute the sum of the given series. If the series diverges, enter DNE. Use exact values. sum from n=0 to infinity of (pi^(2n)) / ((-4)^n(2n)!) =

Solution Steps

To determine the sum of the given series, we need to recognize the pattern and see if it matches a known series. The series resembles the Maclaurin series expansion for the cosine function, which is given by:

\[ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \]

In the given series, the term \(\frac{\pi^{2n}}{(-4)^n(2n)!}\) can be rewritten as \(\frac{(\pi^2/4)^n}{(2n)!}\), which matches the cosine series with \(x = \frac{\pi}{2}\). Therefore, the sum of the series is \(\cos\left(\frac{\pi}{2}\right)\).

The given series is

\[ \sum_{n=0}^{\infty} \frac{\pi^{2 n}}{(-4)^{n}(2 n)!} \]

This series resembles the Maclaurin series expansion for the cosine function, which is

\[ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \]

We can rewrite the terms of the series as follows:

\[ \frac{\pi^{2 n}}{(-4)^{n}(2 n)!} = \frac{(\pi^2/4)^{n}}{(2n)!} \]

This indicates that the series can be expressed in terms of the cosine function with \(x = \frac{\pi}{2}\).

Using the known value of the cosine function:

\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

Thus, the sum of the series converges to \(0\).

Final Answer