So I'm going through the notes for my online summer calculus class, and something struck me as odd about the ratio test: it used variables only, there were no numbers plugged in. For example, in the series $$\sum_{n=1}^\infty\frac{n}{4^n}$$ we have $a_{n+1}=\frac{n+1}{4^{n+1}}$ and $a_n=\frac{n}{4^n}$ as the numerator and denominator in the ratio $ \left | \frac{a_{n+1}}{a_n} \right | $.

Now, my notes say that we then plug $a_{n+1}=\frac{n+1}{4^{n+1}}$ and $a_{n}=\frac{n}{4^n}$ directly into the ratio, giving us

$$ \left | \frac{\frac{n+1}{4^{n+1}}}{\frac{n}{4^n}} \right | $$

which we simplify down to $$ \left | \frac{n+1}{4n} \right | $$

That ratio is then plugged into the limit and which we find is $\frac{1}{4}$, so we know that the series converges since $\frac{1}{4} < 1$.

My question is, why do we solve for the initial ratio algebraically with symbols? Couldn't we just pick an integer $n$ and plug that as well as the integer $n+1$ into the equation to get our ratio? It seems like a lot of extra work that can be simplified.

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    $\begingroup$ Your expression should simplify to ${n+1\over 4n}$. You then take the limit of this expression as $n$ tends to infinity (the limit is $1/4$, here). So, no, you can't simply plug in a number. $\endgroup$ Jul 16, 2012 at 20:21
  • $\begingroup$ Whoops, you caught my typo. I was copying and pasting previous expression so I wouldn't have to rewrite entire expressions, and I forgot to change that one. $\endgroup$ Jul 16, 2012 at 20:25
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    $\begingroup$ If you have uncertainty about what the limit of the ratio (in this case the ratio is $\frac{n+1}{4n}$) might be, it can be informative to use the calculator to find the ratio for some specific large $n$. That does not do the job of calculating the limit of the ratio, but it might lead to an insight about that limit. $\endgroup$ Jul 16, 2012 at 20:27

1 Answer 1


No, you can't just plug in numbers.

The ratio test is supposed to be about $\lim \dfrac{|a_{n+1}|}{|a_{n}|}$, and this limit is the key part.


Let's look into this a bit more. Suppose I have the series whose first thousand terms $a_1, \ldots, a_{1000}$ are $1,2, \ldots, 1000$, i.e. $a_n = n$. But after those terms, the series becomes $a_n = \dfrac{1}{2^n + n}$. If you were to just 'find the ratio' of any two of the first 1000 terms, you would think the series diverges, since you'd get a number that's greater than $1$. And if you were to find the ratio of any two of the subsequent terms, you'd get different ratios for every pair! But the limiting ratio is nonetheless $1/2$, which explains why $\sum a_n$ converges in the end.

  • $\begingroup$ Why wouldn't plugging in numbers into the ratio, then doing $lim\left | ratio \right |$ work? $\endgroup$ Jul 16, 2012 at 20:26
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    $\begingroup$ @Nick How do you take the limit of a number? $\endgroup$
    – davidlowryduda
    Jul 16, 2012 at 20:26
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    $\begingroup$ Once you have chosen $n$, you have "frozen" the ratio, it can't move any more, so you cannot find the limit. Just like if you are trying to find the limit as $h$ approaches $0$ of $\frac{(x+h)^2-x^2}{h}$, and you let $h=1/10$. $\endgroup$ Jul 16, 2012 at 20:31
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    $\begingroup$ It suddenly hit me after I read $1,2,...,1000$ that the ratio of $\frac{3}{2} \neq \frac{1000}{999}$. See, this is why I should not be doing math in the middle of July. Must be the heat. $\endgroup$ Jul 16, 2012 at 20:49
  • $\begingroup$ @mixedmath That's also a good point. The limit of a number is itself and once again, $\frac{3}{2} \neq \frac{1000}{999}$ so the limit of two different ratios would not be the same. $\endgroup$ Jul 16, 2012 at 20:52

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