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# 数学代写|实分析代写Real Analysis代考|The Convergence Test

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## 数学代写|实分析代写Real Analysis代考|The Convergence Test

As it stands, Abel’s lemma does not seem to be much help. The partial sums of the $a$ ‘s in our example never exceed $1.112233 \ldots$, but since $b_1=1$, Abel’s lemma only tells us that for every odd integer $n$,
\begin{aligned} & \mid \cos (0.15 \pi)-\frac{1}{3} \cos (0.45 \pi)+\frac{1}{5} \cos (0.75 \pi)-\frac{1}{7} \cos (1.05 \pi) \ & +\cdots \pm \frac{1}{n} \cos (0.15 n \pi) \mid \leq 1.112234 \end{aligned}
That may be nice to know, but it does not prove convergence.
Abel proved his lemma in order to answer questions about the convergence of power series. His paper of 1826 was the first fully rigorous treatment of the binomial series for all values (real and complex) of $x$. We are almost to a convergence test for Fourier series, but it was Dirichlet who was the first to publicly point out how to pass from Abel’s lemma to the convergence test that we shall apply to Fourier series.

The key is to use the Cauchy criterion. A series converges if and only if the partial sums can be brought arbitrarily close together by taking sufficient terms. The difference between two partial sums is simply a partial sum that starts much farther out. We observe that
\begin{aligned} \left|\sum_{k=m+1}^n a_k\right| & =\left|\sum_{k=1}^n a_k-\sum_{k=1}^m a_k\right| \ & \leq\left|S_n\right|+\left|S_m\right| \ & \leq 2 M . \end{aligned}
If $T_n=\sum_{k=1}^n a_k b_k$ and the $a$ ‘s and $b$ ‘s satisfy the conditions of Abel’s lemma, then
$$T_n-T_m=\sum_{k=m+1}^n a_k b_k \leq 2 M b_{m+1}$$

## 数学代写|实分析代写Real Analysis代考|A Trigonometric Identity

We have proven that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{2 k-1} \cos \left(\frac{(2 k-1) \pi x}{2}\right)$$
converges when $x=0.3$. What about other values of $x$ ? In order to apply Dirichlet’s test, we must prove that once we have chosen $x$, the absolute value of the partial sum of the $a$ ‘s,
$$\left|\sum_{k=1}^n(-1)^{k-1} \cos \left(\frac{(2 k-1) \pi x}{2}\right)\right| \text {, }$$
stays bounded for all $n$.

We let $y$ stand for $\pi x / 2$. We would like to find a trigonometric identity that enables us to simplify
$$\cos y-\cos 3 y+\cos 5 y-\cdots-(-1)^n \cos (2 n-1) y .$$
Such an identity can be found by using the fact that
$$\cos A+i \sin A=e^{i A}$$
$$i \sin y-i \sin 3 y+i \sin 5 y-\cdots-(-1)^n i \sin (2 n-1) y$$
to our summation, we can rewrite it as a finite geometric series which we know how to sum:
\begin{aligned} & {[\cos y+i \sin y]-[\cos 3 y+i \sin 3 y]+[\cos 5 y+i \sin 5 y]-\cdots} \ & \quad-(-1)^n[\cos (2 n-1) y+i \sin (2 n-1) y] \ & \quad=e^{i y}-e^{3 i y}+e^{5 i y}-\cdots-(-1)^n e^{(2 n-1) i y} \ & \quad=e^{i y}\left(1+z+z^2+\cdots+z^{n-1}\right) \ & \quad=e^{i y} \frac{1-z^n}{1-z} \end{aligned}
where $z=-e^{2 i y}$.

## 数学代写|实分析代写Real Analysis代考|The Convergence Test

$1.112233 \ldots$ ，但是由于 $b_1=1$, Abel 引理只告诉我们对于每个奇数 $n$,
$$\left|\cos (0.15 \pi)-\frac{1}{3} \cos (0.45 \pi)+\frac{1}{5} \cos (0.75 \pi)-\frac{1}{7} \cos (1.05 \pi) \quad+\cdots \pm \frac{1}{n} \cos (0.15 n \pi)\right| \leq 1.112234$$

$$\left|\sum_{k=m+1}^n a_k\right|=\left|\sum_{k=1}^n a_k-\sum_{k=1}^m a_k\right| \quad \leq\left|S_n\right|+\left|S_m\right| \leq 2 M .$$

$$T_n-T_m=\sum_{k=m+1}^n a_k b_k \leq 2 M b_{m+1}$$

## 数学代写|实分析代写Real Analysis代考|A Trigonometric Identity

$$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{2 k-1} \cos \left(\frac{(2 k-1) \pi x}{2}\right)$$

$$\left|\sum_{k=1}^n(-1)^{k-1} \cos \left(\frac{(2 k-1) \pi x}{2}\right)\right|,$$

$$\cos y-\cos 3 y+\cos 5 y-\cdots-(-1)^n \cos (2 n-1) y$$

$$\cos A+i \sin A=e^{i A}$$

$$i \sin y-i \sin 3 y+i \sin 5 y-\cdots-(-1)^n i \sin (2 n-1) y$$

$$[\cos y+i \sin y]-[\cos 3 y+i \sin 3 y]+[\cos 5 y+i \sin 5 y]-\cdots \quad-(-1)^n[\cos (2 n-1) y+i \sin (2 n-1) y] \quad=e$$

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