# If $\sum a_n$ diverges and $\lambda_n \to \infty$, does the series $\sum \lambda_na_n$ diverge?

Suppose that the series $\displaystyle \sum a_n$ diverges and $\lambda_n \to \infty$. Does the series $\displaystyle \sum \lambda_na_n$ diverge? And what happens if $\{\lambda_n\}$ is an unbounded increasing sequence?

If $\displaystyle \sum a_n$ is a series of positive terms,then from $\lambda_n \to \infty$ we can say that for any large $M>0$ we can always find $m\in \Bbb{N}$ st $\lambda_n>M \quad \forall n>m$ so that $a_n\lambda_n>a_nM$ therefore the series $\displaystyle \sum a_n\lambda_n$ diverges by comparison test. But what happens if I don't assume it to be positive term series? And why is second part of problem different from first?

• I think you meant to say "$a_n\lambda_n > a_nM$" – IAmNoOne Aug 10 '14 at 5:48
• If you don't assume $a_n > 0$ , than how did you come up with the inequality $a_n \lambda_n > a_n$? – IAmNoOne Aug 10 '14 at 5:54
• yes edited, that's what I'm asking I'm able to do this for $a_n>0$ ,but what happens if I don't assume that? – Bhauryal Aug 10 '14 at 5:56
• I missed your last question. There is no difference, see proofwiki.org/wiki/… – IAmNoOne Aug 10 '14 at 5:57

Let $$\sum a_n=1-{1\over4}+{1\over3}-{1\over8}+{1\over5}-{1\over12}+\cdots$$ alternate the reciprocals of odd numbers and multiples of $4$ and let

$$\lambda_n=\begin{cases} \sqrt n\quad\text{if n is odd}\\ 2\sqrt n\quad\text{if n is even}\\ \end{cases}$$

Then

$$\sum\lambda_n a_n=1-{1\over\sqrt2}+{1\over\sqrt3}-{1\over\sqrt4}+{1\over\sqrt5}-{1\over\sqrt6}+\cdots$$

which is a (conditionally) convergent series (because it's alternating a strictly decreasing sequence).

To see why $\sum a_n$ diverges, suppose it converged. Then we would have

$$\sum a_n-\left(1-{1\over2}+{1\over3}-{1\over4}+{1\over5}-{1\over6}+\cdots \right)={1\over4}+{1\over8}+{1\over12}+\cdots$$

But that last sum diverges.