To prove that $\sum_{n=1}^{\infty} a_{n} = \infty \Rightarrow \sum_{n=2}^{\infty} \frac{a_{n}}{a_1 + a_2 + \cdots + a_{n-1}} = \infty$ we'll analyze the behavior of the terms on the left and the sum on the right to show that the second series diverges if the first does.

Step 1: Define the Partial Sums

Let $S_n = a_1 + a_2 + \cdots + a_n$, which represents the partial sum of the sequence $\{a_n\}$. Since we know that $\sum_{n=1}^{\infty} a_n = \infty$, we have $\lim_{n \to \infty} S_n = \infty.$ This indicates that the sequence $S_n$ is unbounded and diverges to infinity.

Step 2: Express the Series in Terms of $S_n$

Consider the series $\sum_{n=2}^{\infty} \frac{a_n}{S_{n-1}}.$ Each term in this series is given by $\frac{a_n}{S_{n-1}}$, where $S_{n-1} = a_1 + a_2 + \cdots + a_{n-1}$. Since $S_{n-1}$ diverges to infinity, the denominator in each term increases without bound. However, this does not imply that the sum converges, as the numerator $a_n$ might still be large enough to counterbalance the growth of $S_{n-1}$.

Step 3: Use a Comparison Argument

To prove the divergence of $\sum_{n=2}^{\infty} \frac{a_n}{S_{n-1}}$, let's examine the series by finding a lower bound for each term $\frac{a_n}{S_{n-1}}$.

Since $S_n \to \infty$ as $n \to \infty$, for any large enough $N$, we can ensure that $S_{n-1} \leq 2 S_n$ for $n > N$. Thus, for large $n$, $\frac{a_n}{S_{n-1}} \geq \frac{a_n}{2 S_n}.$ This gives us the inequality $\sum_{n=2}^{\infty} \frac{a_n}{S_{n-1}} \geq \frac{1}{2} \sum_{n=2}^{\infty} \frac{a_n}{S_n}.$ To show that $\sum_{n=2}^{\infty} \frac{a_n}{S_n}$ diverges, consider the partial sums of this series.

Step 4: Conclude that the Series Diverges

Since $\sum_{n=1}^{\infty} a_n = \infty$, the terms $\frac{a_n}{S_n}$ cannot decay too quickly. Specifically, if $\sum_{n=2}^{\infty} \frac{a_n}{S_n}$ were to converge, then it would imply that $a_n$ grows slower than $S_n$, contradicting the divergence of $\sum a_n$. Therefore, $\sum_{n=2}^{\infty} \frac{a_n}{S_n} = \infty.$ Thus, $\sum_{n=2}^{\infty} \frac{a_n}{S_{n-1}} = \infty,$ which completes the proof.

Let me know if you have questions or want further explanation!

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1. How does the divergence of $S_n$ help ensure the second series diverges?
2. What role does the comparison inequality play in the proof?
3. Why is it crucial that $S_n$ grows without bound?
4. How does the behavior of $\frac{a_n}{S_n}$ relate to the original series' divergence?
5. Could this result change if $a_n$ were not nonnegative?

Tip: In proofs involving series, comparisons to partial sums or bounding terms often help establish convergence or divergence effectively.