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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,
$\lim w_n=1$

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(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 $n=1,2,...\;$.

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  • $\begingroup$ @JB, thank you very much. +1 for you $\endgroup$
    – MTMA
    Mar 28, 2015 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, 2015 at 19:05
  • $\begingroup$ @MTMA: See my revised answer. $\endgroup$ Mar 28, 2015 at 22:35

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