1
$\begingroup$

Statement of Kummer's test :
If $\displaystyle\sum_{n=1}^{\infty}a_n$ and $\;\displaystyle\sum_{n=1}^{\infty}\frac{1}{b_n}$ are two infinite series of positive real numbers such that $\;\displaystyle\sum_{n=1}^{\infty}\frac{1}{b_n}$ is divergent, then $$\displaystyle\lim_{n\rightarrow \infty} w_n=\begin{cases} l>0, & \text{convergent}\\ l <0, & \text{divergent}\\ l =0, & \text{inconclusive} \end{cases} $$ where $\; w_n=\left(\frac{a_n}{a_{n+1}}b_n-b_{n+1}\right).$

Let us consider $\displaystyle\sum_{n=1}^{\infty}a_n=\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ and $\;\displaystyle\sum_{n=1}^{\infty}\frac{1}{b_n}=\;\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$.

Then $\; w_n=\left(\frac{a_n}{a_{n+1}}b_n-b_{n+1}\right)=\frac{n+1}{n}\rightarrow 1 $ as $n\rightarrow \infty$.

Now the $p$ series($p=2$), $\;\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ is convergent and $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent.

Give an example, where $\displaystyle\sum_{n=1}^{\infty}\frac{1}{b_n}$ is divergent and $\displaystyle\sum_{n=1}^{\infty}a_n$ is also divergent but,
$\lim w_n=1$

$\endgroup$
3
$\begingroup$

(Edited) An example where $\displaystyle\sum_{n=1}^{\infty}\frac{1}{b_n}$ is divergent and $\displaystyle\sum_{n=1}^{\infty}a_n$ is also divergent, but $\lim_{n\rightarrow\infty} w_n=1$, is not possible if $b_n>0\;\;(n=1,2,...),$ since this would violate Kummer's test. An example is possible if the condition $b_n>0\;\;(n=1,2,...)$, which is a requirement of Kummer's test, is dropped. For instance, let $a_n=1$ and $b_n=-n$ for $n=1,2,...\;$.

$\endgroup$
3
  • $\begingroup$ @JB, thank you very much. +1 for you $\endgroup$ – MTMA Mar 28 '15 at 17:14
  • $\begingroup$ How can I choose $\{b_n\}_n$ such that $b_n>0$ satisfying $\lim w_n=1$. $\endgroup$ – MTMA Mar 28 '15 at 19:05
  • $\begingroup$ @MTMA: See my revised answer. $\endgroup$ – John Bentin Mar 28 '15 at 22:35

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.