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Chapter 12: Problem 35

Determine whether the series converges or diverse. $$\sum \frac{2 k}{(2 k) !}$$

Short Answer

Expert verified
The series \(\sum \frac{2k}{(2k)!}\) converges because the limit of the ratio of consecutive terms, \(\lim_{k\to\infty} \frac{2(k+1)}{(2k+2)(2k+1)}\), is \(\frac{1}{2}\) which is less than 1.

Step by step solution

01

State the Ratio Test Rule

According to the Ratio Test, given a series \(\sum a_n\), we should calculate the limit \(\lim_{n\to\infty} \frac{a_{n+1}}{a_n}\). If the limit is: - Less than 1, the series converges, - Greater than 1, the series diverges, - Equal to 1, the test is inconclusive and we need another test.
02

State the given series and find the general term a_n

The given series is \(\sum \frac{2k}{(2k)!}\). Here, the general term \(a_k\) is \(\frac{2k}{(2k)!}\).
03

Find a_{k+1}

To find the term \(a_{k+1}\), we replace k with k+1. So, \(a_{k+1} = \frac{2(k+1)}{(2(k+1))!}\).
04

Calculate the ratio \(\frac{a_{k+1}}{a_k}\)

Now, we need to find the ratio \(\frac{a_{k+1}}{a_k} = \frac{\frac{2(k+1)}{(2(k+1))!}}{\frac{2k}{(2k)!}}\).
05

Simplify the ratio obtained in step 4

To simplify the ratio, we multiply the numerator and denominator by \((2k)!(2(k+1))!\), which gives: \(\frac{a_{k+1}}{a_k} = \frac{2(k+1)(2k)!}{2k(2(k+1))!}\) Now, observe that \((2(k+1))! = (2k+2)(2k+1)(2k)!\). After substituting this into the above expression, we get: \(\frac{a_{k+1}}{a_k} = \frac{2(k+1)}{(2k+2)(2k+1)}\)
06

Find the limit of the ratio \(\frac{a_{k+1}}{a_k}\) as k approaches infinity

We need to find the limit: \[\lim_{k\to\infty} \frac{2(k+1)}{(2k+2)(2k+1)}\] After simplification, we notice that the degree of the numerator and the denominator are the same, and thus we can use the rule that states that the limit is equal to the ratio of the leading coefficients: \[\lim_{k\to\infty} \frac{2}{(2)(2)} = \frac{1}{2}\]
07

Conclusion

Since the limit we found in step 6 is less than 1, according to the Ratio Test, the given series \(\sum \frac{2k}{(2k)!}\) converges.

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