1
$\begingroup$

If $a_n > 0$ and $\lim na_n = \ell$ with $\ell \ne 0$, then the series $\sum a_n$ diverges.

I've tried proving this using the root and ratio tests but to no avail.

$\endgroup$
5
$\begingroup$

Since $\lim\limits_{n\to\infty}na_{n}=l$ so there is $N$ such that for $n>N$ you have $$|na_{n}-l|<\epsilon$$ This says $$na_{n}>l-\epsilon$$ choose your $\epsilon=\frac{l}{2}$ and you have $na_{n}>\frac{l}{2}$ for $n>N$. Now note that $\sum_{n>N}\frac{l}{2n}$ diverges and use comparison test.

$\endgroup$

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.