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Suppose that $(a_n)$ and $(b_n)$ be sequences such that $\lim (a_n)=0$ and $\displaystyle \lim \left( \sum_{i=1}^n b_i \right)$ exists. Define $c_n = a_1 b_n + a_2 b_{n-1} + \dots + a_n b_1$. Prove that $\lim (c_n)=0$

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What have you tried? –  Eckhard Apr 7 '13 at 15:31
    
Is it not $lim( \sum_1^n a_i=0)$? –  Inceptio Apr 7 '13 at 15:43
1  
@Inceptio, that is another (easier?) question. –  vonbrand Apr 7 '13 at 15:54
    
@vonbrand: No. Was just trying use $AM-GM$ relation so that I could find bounds. –  Inceptio Apr 7 '13 at 15:57

5 Answers 5

Counter example:if $a_n,b_n=\frac{(-1)^n}{\sqrt{n}}$,Then $\lim_{n\to \infty}a_n=0$ and $\sum_{i=1}^n\frac{(-1)^i}{\sqrt{i}}$ is convergent (Leibniz test) and we have $c_n=\sum_{k=1}^na_kb_{n-k+1}=(-1)^{n+1}\sum_{k=1}^n\frac{1}{\sqrt{(n-k+1)(k)}}$,now we consider $k(n-k+1)\le(n-k+1)(k+1)=(\frac{n}{2}+1)^2-(\frac{n}{2}-k)^2\le(\frac{n}{2}+1)^2$ and $\mid c_n \mid\ge\frac{2(n+1)}{n+2}$, hence $c_n$ isn't convergent to $0$.

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Suppose moreover that $\displaystyle \sum\limits_{k \geq 1} |b_k|<+ \infty$.

Let $\epsilon>0$ and $N \geq 0$ such that $\displaystyle \sum\limits_{k \geq N } |b_k| <\epsilon$. Also, let $M>0$ be such that $|a_k|<M$ for all $k \geq 0$.

Then $$\left| \sum\limits_{k=1}^n a_{n-k+1}b_k \right| \leq M \sum\limits_{k=N}^n |b_k| + \max\limits_{k \geq n-N+2} |a_k| \cdot \sum\limits_{k=1}^{N-1} |b_k|$$

For $n$ large enough, you get $$\left| \sum\limits_{k=1}^n a_{n-k+1}b_k \right| \leq (M+1)\epsilon$$

Therefore, it is true that $\lim\limits_{n\to + \infty} c_n=0$ when $\displaystyle \sum\limits_{n \geq 1} b_n$ is absolutely convergent.

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Suppose $b_i \geq 0$

I will give something of a hint.

Two observations will lead you to the answer.

$1)$ Since $\lim a_n=0$, $|a_n|$ is bounded

$2)$ The tail of $\sum _{n=0}^{\infty}b_n$ tend to zero. Which mean for $\epsilon >0$ you can find $n_0$ such that $\sum_{n=k}^{m}b_n<\epsilon \forall k,m \geq n_0$.

One good way to start is to use the condition $\lim a_n=0$ and try to make the most out of it. Try to find what part of $c_n$ is pushed to zero by $a_n$. When you find one the role of $a_n$ the other part should go to zero due to the conditions on $b_n$.

Of course this is a process, maybe the part you found is sent to zero by $a_n$ is not the optimal, which could imly the the rest of the $c_n$ cannot be sent to zero by $b_n$. Then you go back and try to increase to role of $a_n$.

Let's start: $\lim b_1 a_n=0$, also $\lim b_2 a_{n-1}=0$ and so on. That shows the right part of $c_n = a_1 b_n + a_2 b_{n-1} + \dots + a_n b_1$ goes to zero. A first good guess of what the role of $a_n$ is following: fix a $k$ then $\lim (b_1 a_n +b_2 a_{n-2}+ \cdots +b_ka_{n-k+1})=0 $

Now use to triangle inequality \begin{align} |c_n |&=|a_1 b_n + a_2 b_{n-1} + \dots + a_n b_1|\\ &\leq |(b_1 a_n +b_2 a_{n-2}+ \cdots +b_ka_{n-k+1})|+|b_{k+1}a_{n-k} + \ldots b_1 a_n|\\ &\leq |(b_1 a_n +b_2 a_{n-2}+ \cdots +b_ka_{n-k+1})|+\sup \{|a_n|\}(\sum_{n=k}^{m}b_n) \end{align}

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You are assuming that $\sum b_i$ converges absolutely, like Seirios. Is this correct? –  robjohn Apr 9 '13 at 15:13
    
@PeterTamaroff: yes. I was just pointing out that this is the same assumption that Seirios made. In fact, Seirios' answer is a bit more inclusive since $b_i\ge0$ implies absolute convergence, but not vice-versa. –  robjohn Apr 10 '13 at 20:30

Suppose $a_n\to0$ monotonically.

Let $B_k=\sum\limits_{j=k}^\infty b_j$. Since the sum of $b_j$ converges, $\bar{B}_n=\sup\limits_{k\ge n}\left|B_k\right|$ decreases monotonically to $0$.

Let $\epsilon\gt0$. Choose $N$ so that if $n\ge N$, $a_{n/2}\bar{B}_0\le\epsilon/4, a_0\bar{B}_{n/2}\le\epsilon/4$ $$ \begin{align} |c_n|& =\left|\sum_{k=0}^na_kb_{n-k}\right|\\ &=\left|\sum_{k=0}^na_k(B_{n-k}-B_{n-k+1})\right|\\ &=\left|a_nB_0-a_0B_{n+1}+\sum_{k=0}^{n-1}a_kB_{n-k}-\sum_{k=0}^{n-1}a_{k+1}B_{n-k}\right|\\ &=\left|a_nB_0-a_0B_{n+1}+\sum_{k=0}^{n-1}(a_k-a_{k+1})B_{n-k}\right|\\ &=\left|a_n\bar{B}_0-a_0\bar{B}_n+\sum_{k=0}^{n/2-1}(a_k-a_{k+1})B_{n-k}+\sum_{k=n/2}^{n-1}(a_k-a_{k+1})B_{n-k}\right|\\ &\le|a_n\bar{B}_0|+|a_0\bar{B}_n|+|a_0\bar{B}_{n/2}|+|a_{n/2}\bar{B}_0|\\ &\le\epsilon \end{align} $$ Thus, $c_n\to0$.

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I am sorry all, the correct one should be $b_n$ is positive. I have got my solution to this problem. WLOG $\displaystyle \sum_{i=1}^{\infty} b_i=1$. Given $\epsilon >0$. Then there exists $M$ such that for all $n>M$, $|a_n|< \epsilon$. Next, there exist $N$ such that for all $n \geq m>N$, $b_m+ \dots b_n < \epsilon/(|a_1|+ \dots |a_N|+1)$. Now, consider $n>M+N$, then $|c_n| \leq (|a_1|b_n+ \dots |a_N|b_{n-N+1})+(|a_{N+1}|b_{n-N}+ \dots |a_n|b_1) \leq \epsilon + \epsilon$. Since $\epsilon$ is arbitrary, $(c_n)$ goes to 0.

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