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I could use hints for this assignment.

Let $(x_n)_{n\in\mathbb{N}}$ and $(a_n)_{k\in\mathbb{N}}$ be (real) sequences with $\lim_{n\to \infty} a_n = a \not= 0$.

Prove that $$\sum_{n=1}^\infty x_n \text{ converges absolutely } \ \iff \sum_{n=1}^\infty a_nx_n\text{ converges absolutely}$$

When I tried proving the "$\Rightarrow$" with the majority criterion I ended up dividing by $x_n$ which should not be well-defined because if the series $\sum_{n=1}^\infty x_n$ converges then $\lim_{n\to \infty} x_n = 0$. I cannot determine any index of a null sequence so that for all following indices $x_n \not= 0$.

Thanks.

Edit: Well, I just noticed that the quotient criterion has $x_n$ in its denominator for $n\ge n_0\in\mathbb{N}$. How do you actually ensure that $x_n \not=0$ for an infinite number of indices?

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  • $\begingroup$ If there were only a finite number of indexes such that $x_n\neq 0$ then you are dealing with a finite sum $\endgroup$ – leo Nov 30 '13 at 3:44
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Since $(a_n)_{n\in\mathbb N}$ converges, it is bounded. Let $M$ be such that $|a_n|\le M$ for all $n$. Then $|a_nx_n|\le M|x_n|$ and $\sum|a_nx_n|\le \sum M|x_n|$. This shows that if $\sum x_n$ converges absolutely, so does $\sum a_nx_n$. In more detail: The partial sums of $\sum|a_n x_n|$ are increasing and bounded above by $\sum M|x_n|=M\sum|x_n|$. It follows that $\sum|a_nx_n|$ exists, since it is the supremum of its partial sums.

Conversely, assume $\sum a_nx_n$ converges absolutely. Here we use that $a_n\to a\ne 0$. This implies that there is an $\epsilon>0$ such that $|a_n|>\epsilon$ for all $n$ sufficiently large (because all the $a_n$ will eventually be closer to $a$ than they will be to $0$, so $|a|-|a_n|\le |a_n-a|<|a|/2$ for all $n$ larger than, say $N_0$, that is, $|a_n|>|a|/2$ for $n>N_0$, and we can take $\epsilon=|a|/2$). Then $\sum_{n>N_0}|a_nx_n|\ge\sum_{n>N_0}|x_n|\epsilon$, or $\sum_{n>N_0}|x_n|\le (1/\epsilon)\sum_{n>N_0}|a_nx_n|\le(1/\epsilon)\sum|a_nx_n|$. Since $\sum_{n\le N_0}|x_n|$ is a finite sum, it follows that $\sum|x_n|=(\sum_{n\le N_0}|x_n|)+(\sum_{n>|N_0|}|x_n|)$ converges.

Note that whether the $x_n$ are $0$ or not was irrelevant for this argument.

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  • $\begingroup$ Thanks for the elaborate answer. - Besides, I was curious if the assertion also works for conditional convergence. $\endgroup$ – Nhat Nov 30 '13 at 15:35
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You cannot ensure that $x_n\neq 0$ for an infinite number of indices, but if $x_n\neq 0$ for only finitely many indices, then you should have no trouble proving that both series converge. Hence, let's move on to the case where $x_n\neq 0$ for infinitely many indices.

If $n_1,n_2,n_3,\ldots$ are all the distinct indices such that $x_{n_k}\neq 0$, listed in increasing order, then $\sum_n x_n$ (respectively, $\sum_n a_nx_n$) converges absolutely if and only if $\sum_k x_{n_k}$ (respectively, $\sum_k a_{n_k}x_{n_k}$) does. Hence to reduce clutter let's focus on the case where $x_n\neq 0$ for all $n$.

In that case, you can directly apply the limit comparison test on the series $\sum_n |x_n|$ and $\sum_n |a_n||x_n|$.

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