How do you test for convergence for #sum sin(n*(pi)/2) / (n!)# for n=1 to infinity?

Answer 1
I'd use comparison. The absolute value of each term is #<= 1/(n!)# and #sum 1/(n!)# converges.

So.the given series converts absolutely, hence it converges.

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Answer 2

You can test for convergence of the series ( \sum_{n=1}^\infty \frac{\sin(n\pi/2)}{n!} ) using the Ratio Test. The Ratio Test states that if ( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ) exists and is less than 1, then the series converges absolutely.

In this case, ( a_n = \frac{\sin(n\pi/2)}{n!} ). So, we need to evaluate:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

If this limit is less than 1, the series converges absolutely. If it's greater than 1 or does not exist, the series diverges.

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Answer 3

You can use the ratio test to test for convergence of the series ( \sum \frac{\sin(n\pi/2)}{n!} ) for ( n = 1 ) to infinity. Apply the ratio test by calculating the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

Where ( a_n = \frac{\sin(n\pi/2)}{n!} ).

Calculate the terms ( a_{n+1} ) and ( a_n ), then find the limit as ( n ) approaches infinity. If the limit is less than 1, the series converges. If it's greater than 1 or undefined, the series diverges.

Let's calculate the terms:

[ a_n = \frac{\sin(n\pi/2)}{n!} ]

[ a_{n+1} = \frac{\sin((n+1)\pi/2)}{(n+1)!} ]

Now, find the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{\sin((n+1)\pi/2)}{(n+1)!}}{\frac{\sin(n\pi/2)}{n!}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{\sin((n+1)\pi/2) \cdot n!}{\sin(n\pi/2) \cdot (n+1)!} \right| ]

[ = \lim_{n \to \infty} \left| \frac{\sin((n+1)\pi/2)}{\sin(n\pi/2)} \cdot \frac{n!}{(n+1)!} \right| ]

[ = \lim_{n \to \infty} \left| \frac{\sin((n+1)\pi/2)}{\sin(n\pi/2)} \cdot \frac{1}{n+1} \right| ]

[ = \lim_{n \to \infty} \left| \frac{\sin((n+1)\pi/2)}{\sin(n\pi/2)} \right| \cdot \lim_{n \to \infty} \frac{1}{n+1} ]

Now, examine the behavior of the sine function as ( n ) approaches infinity. The sine function oscillates between -1 and 1 as ( n ) increases, so ( \frac{\sin((n+1)\pi/2)}{\sin(n\pi/2)} ) will not converge to a finite limit. Therefore, the limit as ( n ) approaches infinity is undefined.

Since the limit is undefined, the ratio test is inconclusive. You might consider using other convergence tests such as the alternating series test or the comparison test.

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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.

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.

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.

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.

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