Question

Prove that if \(\sum a_{k}\) diverges, then \(\sum c a_{k}\) also diverges, where \(c \neq 0\) is a constant.

Short answer

Question: Prove that if a series \(\sum a_k\) diverges, then the series \(\sum c a_k\) also diverges, where \(c\) is a non-zero constant. Answer: Using the definition of a divergent series, we showed that since \(\lim_{n\to\infty} S_n\) (the limit of partial sums of \(\sum a_k\)) does not exist, then the limit of the partial sums \(\lim_{n\to\infty} T_n\) of the series \(\sum c a_k\) also does not exist. This is because \(T_n = c S_n\), and if \(\lim_{n\to\infty} S_n\) does not exist, then \(c \lim_{n\to\infty} S_n = cL\) also does not exist. Therefore, the series \(\sum c a_k\) is also divergent.

Step-by-step solution

Definition of Divergent Series

A series \(\sum a_k\) is said to be divergent if the limit of the partial sums does not exist, or in other words, if \(\lim_{n\to\infty} S_n\) does not exist, where \(S_n = a_1 + a_2 + \dots + a_n\). Now let's consider the series \(\sum c a_k\) and its partial sums \(T_n = c a_1 + c a_2 + \dots + c a_n\). We will now show that the limit of \(T_n\) also does not exist.

Factor c from Partial Sums

First, let's factor out the constant \(c\) from the partial sums of the series \(\sum c a_k\). We have: \(T_n = c a_1 + c a_2 + \dots + c a_n = c (a_1 + a_2 + \dots + a_n) = c S_n\)

Use Limit Properties

Now, recall that if \(L = \lim_{n\to\infty} S_n\) and \(M = \lim_{n\to\infty} T_n\), then we have: \(M = \lim_{n\to\infty} T_n = \lim_{n\to\infty} c S_n\) We know that the series \(\sum a_k\) is divergent, which means by definition that \(L = \lim_{n\to\infty} S_n\) does not exist. Since \(c \neq 0\), we can apply the limit properties to find: \(M = c \lim_{n\to\infty} S_n = cL\) Since \(L\) does not exist, \(cL\) also does not exist. Therefore, the limit \(M = \lim_{n\to\infty} T_n\) does not exist.

Conclusion

Since the limit of the partial sums of the series \(\sum c a_k\) does not exist, we can conclude that this series is also divergent. We have proved that if \(\sum a_k\) diverges, then \(\sum c a_k\) also diverges, where \(c \neq 0\) is a constant.