Prove that the series is divergent for $\lambda \leq 1.$

Question

Consider the series whose general term is as follows: $$u_n=\frac{a_n}{(S_n)^\lambda}$$ with the condition $S_n = \sum_{k=1}^{n}a_k$ and constraints that $0\leq a_n\leq 1,$ with $\{a_n\}$ being a decreasing sequence, $S_n$ is a divergent series and $\lambda \leq 1.$ Show that the series is divergent.

I tried to use the fact that $S_n\leq n$ to get the following lower bound: $$u_n\geq \frac{a_n}{n^\lambda} $$ but this does not get me anywhere. Refer to this similar question for $\lambda>1.$

Edit: As suggested in the comments I tried to show that $$\frac{a_n}{S_n^\lambda}\geq \frac{a_n}{S_n}$$ but this would require to show that $S_n^\lambda \leq S_n$ which I am not sure how to prove. Any ideas?

Answer 1

First show that for $\lambda = 1$, the series $\sum \frac{a_n}{S_n}$ is divergent. As suggested in comments there are many ways to do this, but I find the following argument to be relatively simple.

Since $a_n$ is nonnegative for all $n$ (and ignoring the trivial case where eventually $a_n = 0$), the partial sums $S_n$ are nondecreasing and

$$\left|\sum_{k=n+1}^m \frac{a_k}{S_k} \right| = \sum_{k=n+1}^m \frac{a_k}{S_k} \geqslant \frac{1}{S_m}\sum_{k=n+1}^m a_k = \frac{S_m - S_ n}{S_m} = 1 - \frac{S_n}{S_m}$$

Since $\sum a_n $ is divergent we have $S_n \to +\infty$. With $n$ fixed, $S_n/S_m \to 0$ as $m \to \infty$, and for sufficiently large $m$ we have $S_n/S_m < 1/2$.

Thus, for all $n \in \mathbb{N}$ there exists $m > n $ such that

$$\left|\sum_{k=n+1}^m \frac{a_k}{S_k} \right| > \frac{1}{2}$$

This violates the Cauchy criterion for convergence and, hence, the series diverges.

For $\lambda > 1$ since $S_n \to +\infty$ we have $S_n > 1$ and $S_n^{\lambda} < S_n$ eventually. From here you can finish (as you suggested) by applying the comparison test to show that $\sum \frac{a_n}{S_n^\lambda}$ diverges.