Q. 88 - Calculus

Question

Let $\sum _{k=1}^{\infty }{a}_k be a convergent series and \sum _{k=1}^{\infty }{b}_k be a divergent series.$ Prove that the series $\sum _{k=1}^{\infty }\left(a_k+b_k\right)$ diverges.

Step-by-Step Solution

Answer

Proof by method of contradiction.

$\sum _{k=1}^{\infty }\left(a_k+b_k\right)$ is a divergent series.

Step 1. Given Information.

$\sum _{k=1}^{\infty }{a}_k$ is a convergent series and $\sum _{k=1}^{\infty }{b}_k$ is a divergent series.

Step 2. Proof by method of contradiction.

Let suppose $\sum _{k=1}^{\infty }\left(a_k+b_k\right)$ be a convergent series.

Now, we know the sum of two convergent series is a convergent series.

So, $\sum _{k=1}^{\infty }\left(a_k+b_k\right) - \sum _{k=1}^{\infty }\left(a_k\right)$ will also be convergent.

Which can be implied as $\sum _{k=1}^{\infty }\left(a_k+b_k-a_k\right) = \sum _{k=1}^{\infty }\left(b_k\right)$ must be convergent.

But it is given that $\sum _{k=1}^{\infty }\left(b_k\right)$ is a divergent series.

So it contradicts and so our assumption is wrong.

Thus the series $\sum _{k=1}^{\infty }\left(a_k+b_k\right)$ is a divergent series.