0
$\begingroup$
  1. An infinite series is convergent if from and after some fixed term, the ratio of each term to the preceding term is numerically less than some quantity which is itself numerically less than unity.
    Let the series beginning from the fixed term be denoted by $$u_1+u_2+u_3+u_4+\dotsb,$$ and let $$\frac{u_2}{u_1}<r, \frac{u_3}{u_2}<r,\frac{u_4}{u_3}<r,\dotsc,$$ where $r<1$.
    Then \begin{align*} &u_1+u_2+u_3+u_4+\dotsb\\ &=u_1\left(1+\frac{u_2}{u_1}+\frac{u_3}{u_2}\cdot\frac{u_2}{u_1}+\frac{u_4}{u_3}\cdot\frac{u_3}{u_2}\cdot\frac{u_2}{u_1}+\dotsb\right)\\ &<u_1(1+r+r^2+r^3+\dotsb); \end{align*} that is, $<\frac{u_1}{1-r}$, since $r<1$.
    Hence the given series is convergent.

Does that mean that the series $1 +\frac{1}{2} + \frac{1}{3} +\dotsb$ should be a convergent one? $$S_n = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dotsb$$ Here, we have \begin{gather} \frac{u_n}{u_{n-1}} = \frac{1/n}{1/(n-1)} = \frac{n-1}{n} = 1-\frac{1}{n}\\ ∴\boxed{\frac{u}{u_{n-1}}<1}\\ ∴\text{Series should be convergent} \end{gather}

$\endgroup$
5
  • 1
    $\begingroup$ As $n\to \infty $ the ratio $\to 1$ ;so do you really think the ratio becomes less than $1$ $\endgroup$
    – Learnmore
    Aug 18 '16 at 3:51
  • 3
    $\begingroup$ The issue isn't whether the ratio is less than $1$, it's a matter of whether you can find a fixed number $r<1$ such that the ratio is bounded by $r$ for all $n$. $\endgroup$
    – Joey Zou
    Aug 18 '16 at 3:53
  • 1
    $\begingroup$ Read the condition carefully: "[...] after some fixed term [...]the ratio of each term to the preceding term is numerically less than some quantity which is itself less than unity." Then the ratio is $1-{1\over n}$ as you say, but each term after some fixed term is not less than any number less than 1, for if you have some candidate $1-\epsilon$ then for $n> 1/\epsilon$ you have that $1-{1\over n} > 1-\epsilon$, so you will always surpass any factor you were trying to find. $\endgroup$
    – Adam Hughes
    Aug 18 '16 at 3:55
  • 1
    $\begingroup$ Indeed. I will point out that this is colloquially referred to as the ratio test and is often phrased so that we look at the limit $\lim\limits_{n\to\infty}|\frac{u_n}{u_{n-1}}|$. In the case that the limit is strictly less than one, then it will converge, however in this case the limit is in fact equal to one (despite approaching $1$ from the left side). In the case that the limit equals one, more careful analysis must be applied. Some cases like $\sum\frac{1}{n}$ will diverge, while others like $\sum\frac{1}{n^2}$ will converge. $\endgroup$
    – JMoravitz
    Aug 18 '16 at 4:03
  • $\begingroup$ Formatting tips here. $\endgroup$
    – Em.
    Aug 18 '16 at 4:28
4
$\begingroup$

You need $\frac{u_n}{u_{n-1}} < r < 1$ for some $r < 1$ and all $n $.

For any $r <1$ we can find $r < \frac {u_n}{u_{n-1}} < 1$. So we can not find an appropriate $r <1$.

So we failed the hypothesis. The ratio test fails.

$\endgroup$
1
$\begingroup$
  • Read the hypothesis of the statement carefully;it states that after some finite number of terms the ratio of each term to preceding term is numerically less than $1$.

  • Can you find that number of terms after which this criteria happens?

  • Typically no as the ratio moves to $1$ as $n $ increases

$\endgroup$

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.