To prove that $\sum_{n=1}^{\infty} a_{n} = \infty$, given that $\{a_n\}$ is a nonnegative sequence and

$\sum_{n=2}^{\infty} \frac{a_n}{a_1 + a_2 + \cdots + a_{n-1}} = \infty,$

we'll argue by contradiction. Assume that $\sum_{n=1}^{\infty} a_n < \infty$. This assumption implies that the series of partial sums $S_n = a_1 + a_2 + \cdots + a_n$ is bounded as $n \to \infty$. Let's go through the steps in detail.

Step 1: Understanding the Consequence of the Bounded Sum

If $\sum_{n=1}^{\infty} a_n < \infty$, then the sequence of partial sums $\{S_n\}$ (where $S_n = a_1 + a_2 + \cdots + a_n$) converges to some finite limit, say $L$. Thus, for large $n$, $S_{n-1} = a_1 + a_2 + \cdots + a_{n-1}$ approaches $L$ as well.

Step 2: Analyzing the Term $\frac{a_n}{S_{n-1}}$

Consider the term $\frac{a_n}{S_{n-1}}$ for large $n$:

• Since $\{a_n\}$ is nonnegative and $\sum_{n=1}^{\infty} a_n$ is finite, $a_n \to 0$ as $n \to \infty$.
• Since $S_{n-1}$ converges to $L > 0$ as $n \to \infty$, for large $n$, $S_{n-1}$ will be close to $L$, which is positive and finite.

Therefore, for large $n$, the term $\frac{a_n}{S_{n-1}}$ can be made arbitrarily small because $a_n \to 0$ while $S_{n-1}$ remains bounded away from zero.

Step 3: Convergence of the Series $\sum_{n=2}^{\infty} \frac{a_n}{S_{n-1}}$

Since $\frac{a_n}{S_{n-1}}$ becomes arbitrarily small as $n$ grows, the series $\sum_{n=2}^{\infty} \frac{a_n}{S_{n-1}}$ must converge. This is because we are summing a sequence that tends to zero and does so sufficiently quickly (due to both $a_n \to 0$ and $S_{n-1}$ being bounded away from zero).

Step 4: Contradiction

The convergence of $\sum_{n=2}^{\infty} \frac{a_n}{S_{n-1}}$ contradicts our given condition that

$\sum_{n=2}^{\infty} \frac{a_n}{S_{n-1}} = \infty.$

Thus, our initial assumption that $\sum_{n=1}^{\infty} a_n < \infty$ must be false.

Conclusion

Therefore, we conclude that

$\sum_{n=1}^{\infty} a_n = \infty.$

This completes the proof.

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Would you like further clarification or more details on any specific part of the proof? Here are some additional questions for deeper understanding:

1. Why does $a_n \to 0$ imply that the sequence of partial sums does not grow indefinitely?
2. Can we use a comparison test here to reinforce the argument?
3. How would this proof change if $a_n$ could take negative values?
4. What if we only know that $a_n$ is positive but not necessarily that $a_n \to 0$?
5. How does the growth rate of $S_{n-1}$ affect the behavior of $\frac{a_n}{S_{n-1}}$?

Tip: For nonnegative series, divergence often depends on comparing the terms to the partial sum itself.