Let $\{a_n\}^{\infty}_{n=1},\{b_n\}^{\infty}_{n=1}$ be sequence. Assume $\lim\limits_{n\to \infty}a_n=L,\ \ \lim\limits_{n\to \infty}b_n=M$, how to prove $\lim\limits_{n\to \infty}(a_n+2b_n)=L+2M$ using the formal definition of limit of sequence?

I remember the formal definition goes as following(not sure for sequence though): $\forall \epsilon\gt 0,\exists N\in\Bbb{R}:x\gt N\implies |a_n-L|\lt \epsilon$

Then we can set $\forall \epsilon\gt 0,\exists N\in\Bbb{R}:x\gt N\implies |a_n-L|\lt \frac12\epsilon$

$\forall \epsilon\gt 0,\exists N\in\Bbb{R}:x\gt N\implies |b_n-M|\lt \frac14\epsilon$

Then $\forall \epsilon\gt 0,\exists N\in\Bbb{R}:x\gt N\implies |2b_n-2M|\lt \frac12\epsilon$

Then $|a_n-L|+|2b_n-2M|\lt \epsilon$.

But I am not sure write a proof, could someone help?

  • $\begingroup$ These should be the formal definitions of the given limits:$$(\forall \epsilon > 0)(\exists N \in \Bbb{R})(\forall n \in \Bbb{N})(n > N \implies \lvert a_n-L \rvert < \epsilon)$$ $$(\forall \epsilon > 0)(\exists N \in \Bbb{R})(\forall n \in \Bbb{N})(n > N \implies \lvert b_n-M \rvert < \epsilon)$$ $\endgroup$ – Noble Mushtak Apr 18 '16 at 14:45

You pretty much got all of the ingredients you need for a proof. Here is how I would put it together.

Let $\epsilon>0$ be arbitrary.

Since $\lim_{n\to\infty}a_{n}=L$, we may choose an $N_{1}$ such that $|a_{n}-L|<\epsilon/2$.

Since $\lim_{n\to\infty}b_{n}=M$, we may choose an $N_{2}$ such that $|b_{n}-M|<\epsilon/4\implies |2b_{m}-2M|<\epsilon/2$.

Let $N=\max\{N_{1},N_{2}\}$. Then, if $n\geq N$, we have

$$ |(a_{n}+2b_{n})-(L-2M)|=|(a_{n}-L)+(2b_{n}-2M)|\leq|a_{n}-L|+|2b_{n}-2M|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. $$


Let $\epsilon\gt0$ be given. Then, there exist $N_1,N_2\in\mathbb N$ such that $$|a_n-L|\lt\epsilon$$ for all $n\ge N_1$ and $$|b_n-M|\lt\frac{\epsilon}2$$ for all $n\ge N_2$ Thus, $$|2b_n-2M|\lt{\epsilon}$$ for all $n\ge N_2$

Adding the two inequalities, $$|a_n-L|+|2b_n-2M|\lt2\epsilon$$ for all $n\ge\max{(N_1,N_2)}$

By the triangle inequality, $$|a_n+2b_n-L-2M|\le|a_n-L|+|2b_n-2M|\lt2\epsilon$$ for all $n\ge\max{(N_1,N_2)}$

Here, $2\epsilon$, like $\epsilon$, is arbitrarily small.

Thus, $$\lim_{n\rightarrow\infty}(a_n+2b_n)=L+2M$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.