Prove Convergence of $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$

[mahler1]

Homework Statement .

Let $\{a_n\}_{n \in \mathbb N}$ a sequence of real numbers such that $lim_{n \to \infty} a_n=0$ and let $b_n=a_n+2a_{n+1}-a_{n+2}$.

Prove that $\sum_{n=1}^{\infty} a_n$ is convergent iff $\sum_{n=1}^{\infty} b_n$ is convergent.

The attempt at a solution.

Honestly, I don't have a clue how to prove this. I know that if $\{a_n\}_{n \in \mathbb N}$ is convergent, then for a given $ε>0$, there exists $N : m,n>N$ (suppose $n>m$) $\implies |\sum_{i=1}^ n a_i -\sum_{i=1}^ m a_i|=|a_{m+1}+...+a_n|<\epsilon$. I've tried to relate this to partial sums of the series $\sum_{n=1}^{\infty} b_n$ but I couldn't conclude anything. With the other implication I am also stuck. And I don't see how to use the fact that $lim_{n \to \infty} a_n=0$.

 

[Dick]

Homework Statement .

Let $\{a_n\}_{n \in \mathbb N}$ a sequence of real numbers such that $lim_{n \to \infty} a_n=0$ and let $b_n=a_n+2a_{n+1}-a_{n+2}$.

Prove that $\sum_{n=1}^{\infty} a_n$ is convergent iff $\sum_{n=1}^{\infty} b_n$ is convergent.

The attempt at a solution.

Honestly, I don't have a clue how to prove this. I know that if $\{a_n\}_{n \in \mathbb N}$ is convergent, then for a given $ε>0$, there exists $N : m,n>N$ (suppose $n>m$) $\implies |\sum_{i=1}^ n a_i -\sum_{i=1}^ m a_i|=|a_{m+1}+...+a_n|<\epsilon$. I've tried to relate this to partial sums of the series $\sum_{n=1}^{\infty} b_n$ but I couldn't conclude anything. With the other implication I am also stuck. And I don't see how to use the fact that $lim_{n \to \infty} a_n=0$.

You really shouldn't be stuck on the forward direction. If $\sum_{n=1}^{\infty} a_n=L$ then what are $\sum_{n=1}^{\infty} a_{n+1}$ and $\sum_{n=1}^{\infty} a_{n+2}$? So it should be pretty easy to guess what $\sum_{n=1}^{\infty} b_n$ is and prove it using partial sums. The reverse direction is the harder one. That's where you need $lim_{n \to \infty} a_n=0$. If not then a counterexample is $a_n=1$ for all n. Then $\sum_{n=1}^{\infty} b_n$ converges but $\sum_{n=1}^{\infty} a_n$ doesn't.

 

[mahler1]

You really shouldn't be stuck on the forward direction. If $\sum_{n=1}^{\infty} a_n=L$ then what are $\sum_{n=1}^{\infty} a_{n+1}$ and $\sum_{n=1}^{\infty} a_{n+2}$? So it should be pretty easy to guess what $\sum_{n=1}^{\infty} b_n$ is and prove it using partial sums. The reverse direction is the harder one. That's where you need $lim_{n \to \infty} a_n=0$. If not then a counterexample is $a_n=1$ for all n. Then $\sum_{n=1}^{\infty} b_n$ converges but $\sum_{n=1}^{\infty} a_n$ doesn't.

It's true, the forward direction wasn't difficult:

Lets prove that $\sum_{n=1}^{\infty} b_n=2L-a_1+a_2$

By hypothesis, $\sum_{n=1}^{\infty} a_n$ is convergent. Call its limit $L$, then given $\epsilon>0$, there exists $N_{\epsilon} : \forall n\geq N_{\epsilon} \implies |\sum_{i=1}^n a_i -L|<\dfrac{\epsilon}{3}$.

Then, for $n\geq N_{\epsilon}$, $|\sum_{i=1}^n b_i -(2L-a_1+a_2)|=|\sum_{i=1}^n a_i + 2\sum_{i=1}^n a_{i+1}-\sum_{i=1}^n a_{i+2}-(2L-a_1+a_2)|$. By an index change of the form $j=i+1$ and $k=i+2$, we have

$|\sum_{i=1}^n a_i + 2\sum_{i=1}^n a_{i+1}-\sum_{i=1}^n a_{i+2}-(2L-a_1+a_2)|=|\sum_{i=1}^n a_i + 2\sum_{j=1}^{n+1} a_j-2a_1-\sum_{k=1}^{n+2} a_k+ a_1+a_2-(2L-a_1+a_2)|=|\sum_{i=1}^n a_i-L + 2\sum_{j=1}^{n+1} a_j-2L+L-\sum_{k=1}^{n+2} a_k|\leq |\sum_{i=1}^n a_i-L|+2|\sum_{j=1}^{n+1} a_j-L|+|\sum_{k=1}^{n+2} a_k-L|<\dfrac{\epsilon}{3}+\dfrac{2\epsilon}{3}+\dfrac{\epsilon}{3}=\dfrac{4\epsilon}{3}$.

This proves $\sum_{n=1}^{\infty} b_n$ is convergent and that its limit is $2L-a_1+a_2$

I haven't retried the other direction. Now I'll put the matter in my hands (brain). Thanks!

 

[Dick]

It's true, the forward direction wasn't difficult:

Lets prove that $\sum_{n=1}^{\infty} b_n=2L$

By hypothesis, $\sum_{n=1}^{\infty} a_n$ is convergent. Call its limit $L$, then given $\epsilon>0$, there exists $N_{\epsilon} : n\geq N_{\epsilon} \implies |\sum_{i=1}^n a_i -L|<\dfrac{\epsilon}{3}$. Note that for $n\geq N_{\epsilon}$, $|\sum_{i=1}^n a_{i+1} -L|<\dfrac{\epsilon}{3}$ and $|\sum_{i=1}^n a_{i+2} -L|<\dfrac{\epsilon}{3}$ (this could be proved by an index change of the form $j=i+1$ and $j=i+2$).

Then, for $n\geq N_{\epsilon}$, $|\sum_{i=1}^n b_i -2L|=|\sum_{i=1}^n a_i -L + 2\sum_{i=1}^n a_{i+1} -2L+L-\sum_{i=1}^n a_{i+2}|\leq |\sum_{i=1}^n a_i -L|+2|\sum_{i=1}^n a_{i+1} -L|+|\sum_{i=1}^n a_{i+2}-L|<\dfrac{4\epsilon}{3}$. This proves $\sum_{n=1}^{\infty} b_n$ is convergent and that its limit is $2L$.

I haven't retried the other direction. Now I'll put the matter in my hands (brain). Thanks!

Sort of, but you can't just reindex series like that without paying some attention to the terms you dropping by doing that. I think $\sum_{n=1}^{\infty} b_n=2L-a_0+a_1$.

 

[mahler1]

Sort of, but you can't just reindex like that without paying some attention to the terms you dropping by doing that. I think $\sum_{n=1}^{\infty} b_n=2L-a_0+a_1$.

You're right, but I think the limit is $2L-a_1+a_2$, I suppose you were considering the first term of the sequence to start at $n=0$ and not $1$. Now I correct my post. By the way, I still don't know how to prove the other implication, could you give me one more hint?

 

[Dick]

You're right, but I think the limit is $2L-a_1+a_2$, I suppose you were considering the first term of the sequence to start at $n=0$ and not $1$. Now I correct my post. By the way, I still don't know how to prove the other implication, could you give me one more hint?

Yes, you are right. I was starting at n=0. Sorry. And I haven't even thought about going the other way yet. But I'm not going to do that until tomorrow. Yawn! That gives you some time to think about it and beat me to the punch.

 

[mahler1]

Yes, you are right. I was starting at n=0. Sorry. And I haven't even thought about going the other way yet. But I'm not going to do that until tomorrow. Yawn! That gives you some time to think about it and beat me to the punch.

Thanks for all the suggestions and corrections. I am going to bed now and see if I have some sort of brain illumination while sleeping.

 

[haruspex]

I still don't know how to prove the other implication, could you give me one more hint?

it's essentially the same. What is the difference between the two sums taken from 1 to r?

 

[mahler1]

it's essentially the same. What is the difference between the two sums taken from 1 to r?

Your question led me to this:

Lets call $s_n=\sum_{k=1}^n a_k$, $t_n=\sum_{k=1}^n b_k$. I'll prove that $\lim_{n \to \infty} 2s_n-t_n=a_1-a_2$. From this, one immediately could deduce that $\{s_n\}_{n \in \mathbb N}$ converges iff $\{t_n\}_{n \in \mathbb N}$ converges.

$2s_n-t_n=2\sum_{k=1}^n a_k-\sum_{k=1}^n b_k=2\sum_{k=1}^n a_k-\sum_{k=1}^n a_k-2\sum_{k=1}^n a_{k+1}+\sum_{k=1}^n a_{k+2}=2\sum_{k=1}^n a_k-\sum_{k=1}^n a_k-2\sum_{k=1}^n a_{k}-2a_{n+1}+2a_1+\sum_{k=1}^n a_{k}+a_{n+2}+a_{n+1}-a_1-a_2=2s_n-s_n-2s_n+s_n+(a_{n+2}-a_{n+1}+a_1-a_2)=a_{n+2}-a_{n+1}+a_1-a_2$.

Then, $\lim_{n \to \infty} 2s_n-t_n=\lim_{n \to \infty} a_{n+2}-a_{n+1}+a_1-a_2=a_1-a_2$.

It follows that $\{s_n\}_{n \in \mathbb N}$ converges iff $\{t_n\}_{n \in \mathbb N}$ converges.

 

[haruspex]

Your question led me to this:

Lets call $s_n=\sum_{k=1}^n a_k$, $t_n=\sum_{k=1}^n b_k$. I'll prove that $\lim_{n \to \infty} 2s_n-t_n=a_1-a_2$. From this, one immediately could deduce that $\{s_n\}_{n \in \mathbb N}$ converges iff $\{t_n\}_{n \in \mathbb N}$ converges.

$2s_n-t_n=2\sum_{k=1}^n a_k-\sum_{k=1}^n b_k=2\sum_{k=1}^n a_k-\sum_{k=1}^n a_k-2\sum_{k=1}^n a_{k+1}+\sum_{k=1}^n a_{k+2}=2\sum_{k=1}^n a_k-\sum_{k=1}^n a_k-2\sum_{k=1}^n a_{k}-2a_{n+1}+2a_1+\sum_{k=1}^n a_{k}+a_{n+2}+a_{n+1}-a_1-a_2=2s_n-s_n-2s_n+s_n+(a_{n+2}-a_{n+1}+a_1-a_2)=a_{n+2}-a_{n+1}+a_1-a_2$.

Then, $\lim_{n \to \infty} 2s_n-t_n=\lim_{n \to \infty} a_{n+2}-a_{n+1}+a_1-a_2=a_1-a_2$.

It follows that $\{s_n\}_{n \in \mathbb N}$ converges iff $\{t_n\}_{n \in \mathbb N}$ converges.

Looks good.