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Given $$ \sum_{n=1}^\infty(a_n+b_n)$$

converges, and $$a_n \to 0$$

prove that $$a_1 + b_1 + a_2 + b_2 + \cdots$$

converges.

The question's intention is to show that $a_n \to 0$ is a sufficient condition for the series to converge.

My try:

using the fact that $ \sum_{n=1}^\infty(a_n+b_n)$ converges, we conclude that $$a_n + b_n \to 0$$

by Cauchy theorem of series convergence.

applying limit arithmetic we conclude that $$b_n \to 0$$

for a given $\epsilon >0$, there exist $N>0$ such that for every $n>N$:

$$|a_n|<\epsilon ~~,~~ |b_n|<\epsilon$$

So trying to show that $$|a_{n+1}+b_{n+1} + \cdots + a_{n+p} + b_{n+p}|< \epsilon_0$$ will do the trick (Cauchy).

With the triangle inequality:

$$|a_{n+1}+b_{n+1} + \cdots + a_{n+p} + b_{n+p}| \le |a_{n+1}|+|b_{n+1}| + \cdots + |a_{n+p}| + |b_{n+p}| \lt 2 p \epsilon$$

and here i am stuck...

Note:

The partial sum $$S_n = \sum_{k=1}^n c_n$$ where $$c_{2n-1} = a_n~~,~~c_{2n}=b_n$$

may not converge. example:

$$a_n = (-1)^n ~~,~~ b_n=(-1)^{n+1}$$

without the parentheses the partial sum doesn't converge.

so $$\sum_{n=1}^\infty (a_n + b_n) = (a_1 + b_1) + (a_2 + b_2) + \cdots \ne a_1 + b_1 + a_2 + b_2 + \cdots$$

Michael Hardy
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john_gayl
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    What you're being asked to prove is exactly the definition of $\sum \limits_{n=1}^\infty(a_n+b_n)$ converging. – Git Gud Jun 01 '14 at 20:14
  • $(1+(-1)) + (1+(-1))+... =0 $ but $1-1+1-1+1-1+1...= ?$ – john_gayl Jun 01 '14 at 20:17
  • Infinite sums written with $\ldots$ are just pictorial abuses of notation to mean something written with $\sum$. The natural interpretation of $a_1+b_1+a_2+b_2+\ldots$ is $\sum \limits_{n=1}^\infty \left(a_n+b_n\right)$, (at least to me that in the natural interpretation). If this is not what you mean, that how do you write $a_1+b_1+a_2+b_2+\ldots$ with $\sum$? – Git Gud Jun 01 '14 at 20:21
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    The point is that you also need to consider the sums starting with a $b_n$ and the sums ending with an $a_n$. Let $c_{2n-1} = a_n$ and $c_{2n} = b_n$. Then show that $\sum_{n=1}^\infty c_n$ converges. – Daniel Fischer Jun 01 '14 at 20:22
  • $\sum_{n=1}^\infty(a_n+b_n)=a_1+b_1+a_2+b_2+\cdots$? – kingW3 Jun 01 '14 at 20:25

2 Answers2

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Let $(T_n)$ be the sequence of partial sums for $\displaystyle\sum_{n=1}^{\infty}(a_n+b_n),\;\;\;$ let $\displaystyle\lim_{n\to\infty}T_n=T,\;\;$and

let $(S_n)$ be the sequence of partial sums for the series $a_1+b_1+a_2+b_2+a_3+b_3+\cdots$.

Then $S_{2n}=T_n$ for each n, so $\displaystyle\lim_{n\to\infty}S_{2n}=T$ and

$\displaystyle\lim_{n\to\infty}S_{2n+1}=\lim_{n\to\infty}(S_{2n}+a_{n+1})=T+0=T$ also.

Therefore $\displaystyle\lim_{n\to\infty}S_{n}=T$, so the series converges.

user84413
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  • From other students in my course it seems to be the right way. but why $S_{2n} = T_n $ formally? How can you apply the associative rule if some of the numbers may be negative? – john_gayl Jun 02 '14 at 19:19
  • It should follow easily by induction. – Git Gud Jun 02 '14 at 19:37
  • I understand now my confusion... if say $a_1=1, b_1=-2, a_2 = -3, b_2=4$ than the sum is $1 + (-2) + (-3) + 4$ and not $1 -2 -3 + 4$. In the first case there is no problem with parenthesis. thank you! – john_gayl Jun 02 '14 at 19:47
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Given $\varepsilon>0$, choose $M$ so large that if $m\ge M$ then $|a_m|<\varepsilon/2$ (possible because $a_n\to0$ as $n\to\infty$) and choose $N$ so large that if $n\ge N$ then $$ |(a_1+b_1)+\cdots+(a_n+b_n)-\text{sum}|<\frac\varepsilon2. \tag 1 $$

If $n\ge\max\{M,N\}$ then \begin{align} & |(a_1+b_1)+\cdots+(a_n+b_n)+a_{n+1}-\text{sum}| \tag 2 \\[6pt] \le {} & |(a_1+b_1)+\cdots+(a_n+b_n)-\text{sum}| + |a_{n+1}| < \frac\varepsilon2+\frac\varepsilon2 = \varepsilon. \end{align}

Every partial sum of the series whose convergence is to be proved is of one of the forms $(1)$ and $(2)$, so all differ from the sum by less than $\varepsilon$ when enough terms are taken.

Michael Hardy
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  • But if you used parenthese, aren't you just "proving" what is given, i.e. that $;\sum (a_n+b_n);$ converges? – DonAntonio Jun 01 '14 at 20:33
  • @DonAntonio : No. What is given is that the partial sums $(a_1+b_1)+\cdots+(a_n+b_n)$ can be made as close to the sum as desired by making $n$ big enough. The proposition to be proved says the same is true of the other series that differs only in that it has more partial sums. The partial sums that it has and that the other one lacks are just the ones in $(2)$. – Michael Hardy Jun 01 '14 at 21:16