Suppose $\sum_{n=1}^{\infty}{x_n} < \infty$,$\sum_{n=1}^{\infty}{|y_n - y_{n+1}|} < \infty$

Question

Suppose $\sum_{n=1}^{\infty}{x_n} < \infty$,$\sum_{n=1}^{\infty}{|y_n - y_{n+1}|} < \infty$. Then prove that $\sum_{n=1}^{\infty}{x_ny_n}$ converges.

Since $\sum_{n=1}^{\infty}{x_n} < \infty$, by divergence test, we have $\lim_n{x_n}=0$

Since $ \sum_{n=1}^{\infty}{|y_n - y_{n+1}|} < \infty $, we have $ \sum_{n=1}^{\infty}{(y_n - y_{n+1})} < \infty $. By divergence test. $\lim_n{(y_n-y_{n+1})}=0$. Hence, we have $|y_n - y_m| \leq |y_n - y_{n-1}| + ... + |y_{m+1}-y_m|<\dfrac{\epsilon}{N}+...+\dfrac{\epsilon}{N}=\epsilon$. Hence, we have $(y_n)$ is a cauchy sequence. Hence, $\sum_{n=1}^{\infty}{x_ny_n}$ converges.

Is this proof correct?

Answer 1

This is a direct consequence of the

Generalized Dirichlet Convergence Test

The standard Dirichlet Test for convergence takes two real sequences $a_n$ and $b_n$ where the $a_n$ have bounded partial sums and the $b_n$ tend to $0$ monotonically. We will derive a slightly more general test using the same ideas.

Let $a_n$ and $b_n$ be two complex sequences and let $$ A_n=\sum_{k=1}^na_k\tag{1} $$ Using the convention that $A_0 = 0$, we have that $a_k = A_k - A_{k-1}$. Therefore, $$ \begin{align} \sum_{k=1}^na_kb_k &=\sum_{k=1}^n(A_k-A_{k-1})b_k\\ &=\sum_{k=1}^nA_kb_k-\sum_{k=0}^{n-1}A_kb_{k+1}\\ &=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})\tag{2} \end{align} $$ Equation $(2)$ leads immediately to the following generalization of the Dirichlet convergence test.

Lemma: Suppose $a_n$ and $b_n$ are two complex sequences and each of the following three conditions is satisfied

The partial sums of the $a_n$ are bounded independently of $n$; i.e. $$ \left|\sum_{k=1}^na_k\right|\le A\tag{3} $$ The $b_n$ are of bounded variation; i.e. $$ \sum_{k=1}^\infty\left|b_k-b_{k+1}\right|\le B\lt\infty\tag{4} $$ The $b_n$ tend to $0$; i.e. $$ \lim_{k\to\infty}b_k=0\tag{5} $$ Then, the series $$ \sum_{k=1}^\infty a_kb_k\tag{6} $$ converges, and the absolute value of its limit is no greater than $AB$.

Notice that condition $(4)$ above replaces the monotonicity in the standard Dirichlet test. In fact, it is not hard to show that any real sequence of bounded variation is the difference of two monotonic sequences. This means that for real sequences, the generalized test is neither stronger nor weaker than the standard Dirichlet test, but a sequence of bounded variation might need to be broken into the difference of two monotonic sequences before having the standard Dirichlet test applied.

This version of the Dirichlet test becomes more useful when checking the convergence of complex series.

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Critique of the proposed answer

Point 1: You had that $$ \sum_{n=1}^\infty|y_n-y_{n+1}|\lt\infty $$ There is no need to remove the absolute values, in fact $$ \sum_{n=1}^\infty(y_n-y_{n+1})\lt\infty $$ leaves the possibility that the sum is unbounded below.

Point 2: You use the fact that $|y_n-y_{n+1}|\to0$ to derive that $y_n$ is a Cauchy sequence. This is not true. Consider $y_n=\sqrt{n}$: $$ \begin{align} |y_n-y_{n+1}| &=\sqrt{n+1}-\sqrt{n}\\ &=\frac1{\sqrt{n+1}+\sqrt{n}}\\[3pt] &\to0 \end{align} $$ however, $y_n$ is not Cauchy.

Answer 2

Let $N\ge2$ then by telescoping $$y_N=y_1-\sum_{n=1}^{N-1}(y_n-y_{n+1})$$ hence the sequence $(y_N)$ is convergent since $\displaystyle \lim_{N\to\infty}\sum_{n=1}^{N-1}(y_n-y_{n+1})=\sum_{n=1}^{\infty}(y_n-y_{n+1})<\infty$ so $(y_n)$ is bounded say by $M>0$ and by $$\inf(Mx_n,-Mx_n)\leq x_ny_n\leq\max(Mx_n,-Mx_n)$$ and $\displaystyle \sum_{n=1}^\infty x_n$ is convergent we have the result.