Answer to Question #159451 in Calculus for Vishal

Question

Let {an}∞n=1 and {bn}∞n=1 be sequences both of which diverge to −∞. Then {an +bn}∞n=1 diverge to −∞ and {an ·bn}∞n=1 diverge to +∞

Answer 1

Here we have $\{a_n\}_{n\geq1}$ and $\{b_n\}_{n\geq1}$ both are divergent sequences, which diverge to $-\infin$ .

• $\{a_n+b_n\}_{n\geq1}$

Let us assume that $\{a_n+b_n\}_{n\geq1}$ converges. Then we know by the property of arithmeticity of sequences, both $\{a_n\}_{n\geq1}$ and $\{b_n\}_{n\geq1}$ converge.

But we already have the fact that $\{a_n\}_{n\geq1}$ and $\{b_n\}_{n\geq1}$ are divergent sequences.

So, we get a contradiction.

Hence our assumption is false, i.e. $\{a_n+b_n\}_{n\geq1}$ diverges.

Also, as $\{a_n\}_{n\geq1}$ and $\{b_n\}_{n\geq1}$ diverge to $-\infin$ , by arithmetic property we have the sequence $\{a_n+b_n\}_{n\geq1}$ diverges to $-\infin$ .

• $\{a_n\cdot b_n\}_{n\geq1}$

Let us assume that $\{a_n\cdot b_n\}_{n\geq1}$ converges. Then we know that $\{a_n\cdot b_n\}_{n\geq1}$ is bounded.

So, $\exist$ c$\in\R$ such that $|a_n\cdot b_n|\leq c$ $\forall$ n.

Since $b_n$ diverges to $-\infin$ , we can get $b_n<0$ for all n.

So, we now have,

$|a_n|=|\frac{a_n\cdot b_n}{b_n}|\leq\frac{c}{|b_n|}$ for all n.

So ,we get that $\{a_n\}_{n\geq1}$ converges to 0, which is a contradiction as we already have $\{a_n\}_{n\geq1}$ diverges to $-\infin$ .

Hence our assumption is false, and $\{a_n\cdot b_n\}_{n\geq1}$ diverges.

Also, as $\{a_n\}_{n\geq1}$ and $\{b_n\}_{n\geq1}$ diverge to $-\infin$ , by arithmetic property we have the sequence $\{a_n\cdot b_n\}_{n\geq1}$ diverges to $\infin$ .