# Infinite limit of the sum of two sequences (one diverges to infinite and the other converges)

We have $$\{a_n\}_{n\in N}$$ $$\{b_n\}_{n\in N}$$ both sequences of real numbers.

I have to prove that:

$$\lim_{n\to\infty} a_n = l \in R , \lim_{n\to\infty} b_n = +\infty \Longrightarrow\lim_{n\to \infty}(a_n +b_n) = +\infty$$

Everything with the limit's definition,

$$\lim_{n\to\infty} a_n = l \Longleftrightarrow \forall\epsilon > 0 , \exists n_o\in N , \forall n>n_0 : |a_n-l|<\epsilon$$

$$\lim_{n\to\infty} b_n = +\infty \Longleftrightarrow \forall M\in R, \exists n_0\in N,\forall n>n_0 : b_n>M$$

What can I do?

I tried writing $$|a_n-l|<\epsilon$$ as $$-\epsilon to get that $$a_n <\epsilon + l$$ and $$a_n> -\epsilon + l$$

Set $$M = M+\epsilon - l$$

Then: $$b_n >M+\epsilon - l$$ and $$a_n> -\epsilon + l$$

Every item on the previous equations are real numbers, which are partially ordered, so we can add those two equations and get:

$$b_n +a_n>M+\epsilon - l -\epsilon +l$$

Which equals to $$b_n +a_n>M$$

## 1 Answer

You're almost there. It seems that your difficulty lies in writing it down cleanly. A proof like this often requires you to "follow" every step in the statement you want to prove, and show that "everything works". Enough vagueness.

In this case, we want to prove that $$\forall M>0:\exists n_0\in\mathbb{N}:\forall n \geqslant n_0: a_n+b_n > M$$. In terms of proving it, we'll want to choose an $$M$$, after which we construct an $$n_0$$ with what we're given, and most likely depending on the $$M$$ we've chosen. Then we show that that $$n_0$$ gives us what we want. In your question you seem to forget about the $$n_0$$, which is probably where you're stuck.

Proof: Choose a real $$M$$, then we can see that there exists an $$n_{0,a} \in \mathbb{N}$$ such that for all $$n\geqslant n_{0,a}$$, $$|a_n-l|<1$$ (The 1 can be an arbitrarily chosen $$\epsilon >0$$). From the last inequality follows that for such $$n$$, $$a_n>-1+l$$.

Likewise, there exists an $$n_{0,b}\in \mathbb{N}$$ such that for all $$n\geqslant n_{0,b}$$, $$b_n>M+1-l$$.

Now, let $$n_0 = \mathrm{max}(n_{0,a},n_{0,b})$$, then for $$n\geqslant n_0$$, we get both $$a_n>-1+l$$ and $$b_n>M+1-l$$. We can add these equalities to get $$a_0+b_0>M$$.