Prove or disprove: if $\{b_{n}\}$ is a convergent sequence and $c_{n} = n(b_{n} - b_{n-1})$, then $\{c_{n}\}$ is a bounded sequence.

Question

The question says:

Prove or disprove: if $\{b_{n}\}$ is a convergent sequence and $c_{n} = n(b_{n} - b_{n-1})$, then $\{c_{n}\}$ is a bounded sequence.

Could you please give me a hint? I am stucked.

Answer 1

Since $$c_n=n(b_n-b_{n-1}),$$ then $$b_n-b_{n-1}=\frac{c_n}{n}.$$ Thus $$b_n-b_0=\sum_{k=1}^n(b_k-b_{k-1})=\sum_{k=1}^{n}\frac{c_k}{k},$$ which implies that $$b_n=b_0+\sum_{k=1}^n \frac{c_k}{k}.$$

Since $\{b_n\}$ is convergent, hence $\sum\limits_{k=1}^{\infty}\dfrac{c_k}{k}$ is also convergent, but $\{c_k\}$ is not necessarily convergent, even not necessarily bounded. Here is a counterexample. Consider $c_k=(-1)^{k}\sqrt{k}$, which is not convergent and not bounded. But the alternating series $$\sum_{k=1}^\infty \frac{c_k}{k}=\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}=(\sqrt{2}-1)\zeta\left(\frac{1}{2}\right)=-0.604899\cdots$$

Answer 2

False. Let $A=\{m^2: m\in \mathbb N\}.$ Define $b_n = 1/\sqrt n$ if $n\in A,$ $b_n=0$ otherwise. Then $b_n\to 0.$ But let's observe what happens to $n(b_n - b_{n-1})$ as $n\to \infty$ within $A.$ For such $n,$

$$n(b_n - b_{n-1}) = m^2(1/m -0) = m\to \infty.$$

Thus $n(b_n - b_{n-1})$ is unbounded.

Answer 3

Let $b_n=0$ when $n$ is even and let $b_n=1/\sqrt n$ when $n$ is odd. Then $b_n$ converges to $0$ but $c_n=\sqrt n$ for all odd $n.$