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# 数学代写|实分析代写Real Analysis代考|Limitations of the Root and Ratio Tests

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## 数学代写|实分析代写Real Analysis代考|Limitations of the Root and Ratio Tests

While the root and ratio tests are usually the ones we want to use first, there are many important series for which they return an inconclusive result. Neither of these tests will confirm that the harmonic series diverges. For the limit ratio test we have
$$\lim {n \rightarrow \infty} \frac{1 /(n+1)}{1 / n}=\lim {n \rightarrow \infty} \frac{n}{n+1}=1$$
Similarly, the limit root test returns
$$\lim {n \rightarrow \infty} n^{-1 / n}=e^{\lim {n \rightarrow \infty}-(\ln n) / n}=e^0=1 .$$
Of course, we know that the harmonic series diverges. We can use this information with the comparison test. If $p \leq 1$, then $1 / n^p \geq 1 / n$ and so $\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges. What if $p$ is greater than 1 ? Does
$$\sum_{n=1}^{\infty} \frac{1}{n^{1.01}}$$
converge or diverge? Can we find a divergent series with $a_n<1 / n$ ? What about $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ ? Our last two tests enable us to answer these questions. They are both based on the observation that if the summands are positive, then the partial sums are increasing. If the partial sums are bounded, then they form a Cauchy sequence and so the series converges. If the partial sums are not bounded, then the series diverges to infinity.

## 数学代写|实分析代写Real Analysis代考|Cauchy’s Condensation Test

The first convergence test in Cauchy’s Cours d’analyse is the root test. The second is the ratio test. The third is the condensation test.

Theorem 4.11 (Cauchy’s Condensation Test). Let $a_1+a_2+a_3+\cdots$ be a series whose summands are eventually positive and decreasing. That is to say, there is a subscript $N$ such that
$$n \geq N \quad \text { implies that } \quad a_n \geq a_{n+1} \geq 0 .$$
This series converges if and only if the series
$$a_1+2 a_2+4 a_4+8 a_8+\cdots+2^k a_{2^k}+\cdots$$
converges.
This test is good enough to settle the convergence questions that the root and ratio tests could not handle. We shall state and prove the $p$-test after we have proven Cauchy’s test. But first, we show that there is a series with smaller summands than the harmonic series but which still diverges. We consider
$$\frac{1}{2 \ln 2}+\frac{1}{3 \ln 3}+\frac{1}{4 \ln 4}+\cdots$$
These summands are positive and decreasing. We can apply the condensation test, letting the first summand be 0 and treating $1 / 2 \ln 2$ as the second summand. We compare our series with
\begin{aligned} \frac{2}{2 \ln 2}+\frac{4}{4 \ln 4}+\frac{8}{8 \ln 8}+\cdots & =\sum_{k=1}^{\infty} \frac{2^k}{2^k \ln 2^k} \ & =\sum_{k=1}^{\infty} \frac{1}{k \ln 2} \ & =\frac{1}{\ln 2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right) \end{aligned}
We are comparing our original series with the harmonic series which we know diverges. It follows that $1 /(2 \ln 2)+1 /(3 \ln 3)+\cdots$ also diverges.

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## 数学代写|实分析代写Real Analysis代考|Limitations of the Root and Ratio Tests

$$\lim n \rightarrow \infty \frac{1 /(n+1)}{1 / n}=\lim n \rightarrow \infty \frac{n}{n+1}=1$$

$$\lim n \rightarrow \infty n^{-1 / n}=e^{\lim n \rightarrow \infty-(\ln n) / n}=e^0=1 .$$

$$\sum_{n=1}^{\infty} \frac{1}{n^{1.01}}$$

## 数学代写|实分析代写Real Analysis代考|Cauchy’s Condensation Test

Cauchy 的 Cours d’analyse 中的第一个收玫检验是根检验。二是比例测试。三是结 露试验。

$$n \geq N \quad \text { implies that } \quad a_n \geq a_{n+1} \geq 0 .$$

$$a_1+2 a_2+4 a_4+8 a_8+\cdots+2^k a_{2^k}+\cdots$$

$$\frac{1}{2 \ln 2}+\frac{1}{3 \ln 3}+\frac{1}{4 \ln 4}+\cdots$$

$$\frac{2}{2 \ln 2}+\frac{4}{4 \ln 4}+\frac{8}{8 \ln 8}+\cdots=\sum_{k=1}^{\infty} \frac{2^k}{2^k \ln 2^k} \quad=\sum_{k=1}^{\infty} \frac{1}{k \ln 2}=\frac{1}{\ln 2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right)$$

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