Prove if a series $\sum{a_n}$ is convergent, then $b_n = \frac{1}{a_{n+1}} - \frac{1}{a_n}$ is unbounded

Question

Let $(a_n)_{n \geq 1}$ be a decreasing sequence of positive reals. Let $s_n = a_1 + a_2 + ... + a_n$ and

\begin{align} b_n = \frac{1}{a_{n+1}} - \frac{1}{a_n}, n \geq 1 \end{align}

Prove that if $(s_n)_{n \geq 1}$ is convergent, then $(b_n)_{n \geq 1}$ is unbounded.

My attempt: If $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} \neq 1 $ or DNE, then we can bound $\frac{a_{n+1}}{a_{n}}$ by $\frac{a_{n+1}}{a_{n}} < \frac{1}{k}$ for some $ k>1$

Hence $b_n = \frac{1}{a_{n+1}} - \frac{1}{a_n} > \frac{k-1}{a_n} \to \infty$

On the other hand, if $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 1$ I'm not sure how to prove for this case.

I tried $\forall \epsilon >0 , \exists N>0, n>N \implies |\frac{a_{n+1}}{a_n} - 1| < \epsilon \implies |\frac{1}{a_{n+1}} - \frac{1}{a_n}| < \frac{\epsilon}{1-\epsilon} \frac{1}{a_n}$.

But this doesn't help me as I am looking for a lower bound.

Any hints are appreciated

Answer 1

This is an old problem but I think a more detailed solution is worth having.

Suppose (1) $0<a_n\searrow0$ and (2) $\sum_na_n<\infty$, and (3) $B:=\sup_n\Big(\frac{1}{a_{n+1}}-\frac{1}{a_n}\Big)<\infty$. By (1) and (2) $\lim_\limits{n\rightarrow\infty} na_n=0$: $$na_{2n}\leq a_{n+1}+\ldots + a_{2n}\xrightarrow{n\rightarrow\infty}0$$ and $$(n+\frac12)a_{2n+1}\leq (n+1)a_{2n+1}\leq a_n+\ldots + a_{2n}\xrightarrow{n\rightarrow\infty}0$$ By (3), \begin{align} n\Big(1-\frac{a_{n+1}}{a_n}\Big)\leq B na_{n+1}\xrightarrow{n\rightarrow\infty}0 \end{align} It follows from Raabe's theorem that $\sum_na_n$ diverges, which contradicts (2). Hence, no such sequence $a_n$ exists.

Answer 2

Hint: Suppose for the sake of contradiction that the $b_n$ were bounded. Then show that $\sum{a_n}$ is divergent.