Solve the problem.
-If $\sum a _ { n }$ is a convergent series of nonnegative terms, what can be said about $\sum n ^ { k } a _ { n }$, where $k$ is a positive integer?
A) Always converges
B) May converge or diverge
C) Always diverges

Question

Quizplus · Accepted Answer

We cannot say for sure whether $\sum n^k a_n$ converges or diverges. However, we can use the comparison test to make a prediction. Since the terms of $\sum a_n$ are nonnegative, we can compare $n^k a_n$ to just $a_n$ and see if $\sum a_n$ converges or diverges. If $k \leq 1$, then $n^k \leq n$ and we can conclude that $\sum n^k a_n$ may diverge if $\sum a_n$ diverges, but it necessarily converges if $\sum a_n$ converges. This is because, in that case, we have the inequality $$0 \leq \sum_{n=1}^{\infty} n^k a_n \leq \sum_{n=1}^{\infty} n a_n$$which converges by the comparison test (since $\sum a_n$ converges). However, if $k > 1$, then $n^k \geq n$ and we cannot make the same prediction as above. In fact, it is possible for $\sum n^k a_n$ to diverge even if $\sum a_n$ converges in this case. For example, consider the series $$a_n = \begin{cases} \frac{1}{n^2} & \text{if } n \text{ is a perfect square} \ \frac{1}{n^3} & \text{otherwise} \end{cases}$$which converges since $$\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{1}{n^3} + \sum_{k=1}^{\infty} \frac{1}{k^2} \cdot \frac{1}{k^3} \leq \zeta(3) + \sum_{k=1}^{\infty} \frac{1}{k^5}$$where $\zeta$ denotes the Riemann zeta function. However, the series $\sum n^2 a_n$ diverges since $$\sum_{n=1}^{\infty} n^2 a_n = \sum_{n=1}^{\infty} \frac{1}{n} + \sum_{k=1}^{\infty} k^2 \cdot \frac{1}{k^3} \geq \sum_{n=1}^{\infty} \frac{1}{n}$$which diverges. Therefore, the correct answer is B.

Solve the Problem ∑an\sum a _ { n }∑an​ Is a Convergent Series of Nonnegative Terms, What Can Be

Solve the Problem $\sum a _ { n }$ Is a Convergent Series of Nonnegative Terms, What Can Be