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Find the radius of convergence and the interval of
convergence
n!x^n/1.3.5......(2n-1)

User Timseal
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1 Answer

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To find the radius and interval of convergence of the series, we employ the ratio test.

The ratio test states that, for a series, if the limit (as n approaches infinity) of the absolute value of the ratio of the (n+1)th term to the nth term of the series is less than 1, the series converges.

For our series n!x^n/1.3.5......(2n-1), we can set up the following equation:

| ( (n+1)!x^(n+1)/1.3.5......(2n+1) ) / ( n!x^n/1.3.5......(2n-1) ) | < 1

Now, simplify this equation. The factorial and power terms will cancel out leaving us:

| (n+1)*x / (2n+1) | < 1

Next, as n tends towards infinity, the ratio modifies to:

| x / 2 | < 1

From this, we establish that the radius of convergence, R, is 2. The radius of convergence is the value that the absolute value of x must be below for the series to converge.

Once we have the radius of convergence, we can determine the interval of convergence. This interval includes all x such that |x| < R. In other words, -R < x < R.

Therefore, given R=2, the interval of convergence I is -2 < x < 2.

In summary, for the series n!x^n/1.3.5......(2n-1), the radius of convergence is 2, and the interval of convergence is -2 < x < 2.

User Oleh Prypin
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