I a newbie to series, and I have not done too much yet. I have an exercise where I have basically to say if some series are convergent or divergent. If convergent, determine (and prove) the sum of the series.

This is the first series:

$$\sum_{i=0}^n \frac{4}{3^i}$$

I have heard about the ratio test, so I decided to try to apply it in this case to see if the series is or not convergent.

The ratio test is basically defined like this:

$$L = \lim{\left|\frac{a_{n+1}}{a_n}\right|}$$

Where $a_n$ is in this case $\frac{4}{3^n}$ and $a_{n+1}$ is consequently $\frac{4}{3^{n+1}}$. Thus, we have:

$$L = \lim{\left|\frac{\frac{4}{3^{n+1}}}{\frac{4}{3^n}}\right|} = \lim{\left|\frac{4 \cdot 3^n}{3^{n+1} \cdot 4}\right|} = \lim{\left|\frac{3^n}{3^{n+1}}\right|} = \lim{\left|\frac{3^n}{3^n\cdot 3}\right|} = \lim{\left|\frac{1}{3}\right|} = \frac{1}{3} < 1$$

From the ratio test, we know if $L< 1$, then the series converges.

Now, I need to find the sum. I have followed a video which explain how to find the sum using the ratio, which sincerely I have not understand well what it is. Can you explain a general rule to find the ratio, and what exactly is it?

Ok. So, following the same process done during the video, I have:

$$\sum_{i=0}^n \frac{4}{3^n} = 4 \sum_{i=0}^n \frac{1}{3^n} = 4 \sum_{i=0}^n \left(\frac{1}{3}\right)^n$$

Apparently, our ratio $r$ is $\frac{1}{3}$.

Now, to find the sum of the series, he uses a formula:

$$\frac{\text{first term}}{1 - r}$$

Where first term is the first term of the series. Can you explain where this formula comes from?

Applying that formula, we obtain:

$$\frac{\left(\frac{1}{3}\right)^0}{1 - \frac{1}{3}} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$$

I checked the sum using a online calculator, and apparently it is not $\frac{3}{2}$, so I decided to manipulate, and I thought I left a $4$ outside the series, so I multiply that $4$ by $\frac{3}{2}$, which produces $\frac{12}{2} = 6$, which is exactly the sum I found using 2 online calculators.

Now, my 3rd question is: why what I did is correct (or not correct)?

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    $\begingroup$ Yes, it is correct. $\endgroup$ – Timbuc Mar 28 '15 at 16:24
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    $\begingroup$ The $\frac{\text{first~term}}{1-r}$ formula comes from $(1-r)(1+r+r^2+\ldots + r^n) = 1-r^{n+1}$ (multiply out to see why). If $|r|<1$ then $r^{n+1} \to 0$ as $n\to \infty$ giving the formula $1+r+r^2 + \ldots = \frac{1}{1-r}$. See for example this and this $\endgroup$ – Winther Mar 28 '15 at 16:27
  • $\begingroup$ mathworld.wolfram.com/GeometricSeries.html $\endgroup$ – Poppy Mar 28 '15 at 16:29
  • $\begingroup$ @Winther So, as far as I understood from the last statement from the WolframMathWorld webpage you sent me, if the sum started from $k=1$, instead of $k=0$, instead of $1$ and the numerator, we could have $r$, the ratio, so we would have the formula $\frac{r}{1 - r}$? $\endgroup$ – nbro Mar 28 '15 at 16:37
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    $\begingroup$ @Rinzler That is correct, and in general if the sum starts at $k=n$ then it's $\frac{r^n}{1-r} = \frac{\text{first term}}{1-r}$. To prove this just multiply $1+r+r^2+\ldots = \frac{1}{1-r}$ by $r^n$ to get $r^n + r^{n+1} + r^{n+2} + \ldots = \frac{r^n}{1-r}$. $\endgroup$ – Winther Mar 28 '15 at 17:06

This will be a very long answer, but infinite sums were one of my favorite things in calculus.

First I must point out that there is a typo in your sum. Notice that $$\sum_{i=0}^n\frac{4}{3^n} = \frac{4}{3^n}+\frac{4}{3^n}+\frac{4}{3^n}+\ldots + \frac{4}{3^n} = (n+1)\frac{4}{3^n}$$ while I believe the geometric sum you are interested in is $$\sum_{i=0}^n\frac{4}{3^i} = \frac{4}{3^0}+\frac{4}{3^1}+\frac{4}{3^2}+\ldots + \frac{4}{3^n}$$

First question: You correctly applied the ratio test and proved that this series will converge at infinity. As for a general rule to find the ratio, I'll do my best to explain with an example. In summary you have to $(1)$ look for a pattern, and $(2)$ then pattern match. Let's say you are looking at a sum $$C+\frac{C}{2}+\frac{C}{4}+\frac{C}{8}+\frac{C}{16}+\ldots$$ where $C$ is some constant. There is nothing wrong with factoring out the $C$ since it is not changing from numerator to numerator (the same cannot be said about the denominator.) So, $$C+\frac{C}{2}+\frac{C}{4}+\frac{C}{8}+\frac{C}{16}+\ldots = C\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots\right)$$ Now we can focus on the quantity inside the parentheses. This is where you need to start looking for patterns. By that I mean notice that every number in the parentheses can be rewritten as $\frac{1}{2^n}$ for some integer $n$. It should be clear that $\frac{1}{2}$ is a common ratio of each term. If this is true of every number in the series, then we have what is called a geometric series. Convince yourself of the following equality: $$C+\frac{C}{2}+\frac{C}{4}+\frac{C}{8}+\frac{C}{16}+\ldots = C\sum_{n=0}^\infty \frac{1}{2^n}$$ Whenever you are given a geometric sum, you want to pattern match it to this form. That is, pull out a constant if you can, and make it so the common ratio is clearly discernible inside the sum, raised to the power of $n$, starting at $n=0$. This will allow you to use the formula to find the value of a convergent geometric series. Sometimes $C$ will also look like the common ratio. This can be a bit confusing. For example, the sum $$4+2+1+\frac{1}{2}+\ldots$$ is identical to the sum above, except that $C=4$. Factoring out a $4$ from the sum will leave you with exactly the same series as above. If you come across the sum $$\sum_{n=1}^\infty \left(\frac{1}{3}\right)^{2n}$$ you will need to do some "pattern matching." First notice that $\left(\frac{1}{3}\right)^{2n} = \left(\frac{1}{9}\right)^{n}$ so $$\sum_{n=1}^\infty \left(\frac{1}{3}\right)^{2n} = \sum_{n=1}^\infty \left(\frac{1}{9}\right)^{n}$$ But now the index doesn't start at $n=0$. So let's factor out $\frac{1}{9}$ from each term. Since $\left(\frac{1}{9}\right)^{n}=\frac{1}{9}\left(\frac{1}{9}\right)^{n-1}$ then we have $$\sum_{n=1}^\infty \left(\frac{1}{9}\right)^{n}=\frac{1}{9}\sum_{n=1}^\infty \left(\frac{1}{9}\right)^{n-1} = \frac{1}{9}\sum_{n=0}^\infty \left(\frac{1}{9}\right)^{n}$$ and have now transformed the sum to one that can easily be dealt with.

More generally, if you have a sum $$x_0+x_1+x_2+x_3+\ldots$$ where $x_i$ is some number and you realize that you can use algebra to rewrite each $x_i$ as $ar^i$ where $a,r$ are constants, then you will have a geometric sum with common ratio $r$ multiplied by $a$. Become friends with the rules of exponents! You will have to use these almost every time you are looking for patterns or pattern matching.

Second question: The formula you mention is pretty easy to derive. Let's define $S_n = a+ar+ar^2+ar^3 + \ldots + ar^n$. It should be clear that this is a geometric sum with common ratio $r$. Now multiply $S_n$ by $r$ to get $rS_n = ar+ar^2+ar^3+\ldots ar^{n+1}$. We will subract $S_n$ from $rS_n$ to solve for $S_n$ so that we can find a different but equivalent form of $S_n$. Sounds kind of weird but it's very useful. $$rS_n-S_n = \left(ar+ar^2+ar^3+\ldots ar^{n+1}\right)-\left(a+ar+ar^2+ar^3 + \ldots + ar^n\right) \\ \implies S_n(r-1) = ar^{n+1}+(ar^n-ar^n)+(ar^{n-1}-ar^{n-1})+\ldots + (ar-ar)-a \\ \implies S_n(r-1) = ar^{n+1}-a \\ \implies S_n = \frac{ar^{n+1}-a}{r-1}$$ Equality holds for all $r\neq 1$. Next we need to do a little bit of analysis on $ar^{n+1}$. If $|r|>1$, it should be clear that $ar^{n+1}$ will get bigger as $n$ increases (or become more negative if $r<-1$). This means the quantity $\frac{ar^{n+1}-a}{r-1}$ will not be well behaved as $n \to \infty$ since $ar^{n+1}$ will grow out of control. Now if $|r|<1$, then $ar^{n+1}$ will get smaller and smaller as $n$ increases. You typically won't prove rigorously in a beginner calculus class that $ar^{n+1} \to 0$ as $n \to \infty$, but that is what will happen. I hope you don't mind me handwaving past this detail as my calc teachers did to me! I can include a rigorous proof if you don't believe me. If you do take my word on this detail, then $$\lim_{n \to \infty} \frac{ar^{n+1}-a}{r-1} =\frac{0-a}{r-1} \\ = \frac{-a}{(-1)(1-r)} \\ = \frac{a}{1-r}$$ which establishes the formula for convergent geometric series (geometric series with common ratio $|r|<1$).

Third question: It might be difficult to answer this question without sounding redundant. You are correct in your work on this geometric series because you correctly pattern matched your series to the form that lets you use the equation $\frac{a}{1-r}$. You correctly followed all the necessary steps that one must go through to evaluate a convergent geometric series. Therefore what you did is right. This is one of those nice things about mathematics. If you do everything right every step of the way, you will always be correct.

  • $\begingroup$ Ok. Thanks for all the explanations. I admit I still have to revise your answer to understand everything, but I do not have much time unfortunately. To find the ratio in the case of my question, it was not so complicated, but what about if you have something like: $\sum_{i=0}^n \frac{n^4}{2^n}$ or $\sum_{i=0}^n \frac{(-1)^k k}{k+2}$? To discover if a series is convergent, in the first case is not so complicated, you do the limit of the absolute value of $\frac{a_{n+1}}{a_n}$, but how do you find for example the ratio in those 2 cases (in order to calculate the sum)? $\endgroup$ – nbro Mar 28 '15 at 17:44
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    $\begingroup$ @Rinzler ah, my answer was definitely focused on geometric series. The notion of "finding a common ratio" applies only to geometric series. The two series in your comment above are not geometric series, so there is no common ratio to be found. To determine if they converge you can still use the ratio test and evaluate $$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$$ Note that the value obtained through the ratio test is not the common ratio, unless your series was geometric to begin with. $\endgroup$ – graydad Mar 28 '15 at 17:50
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    $\begingroup$ As for applying the ratio test to the two series you mentioned, we have $$\lim_{n \to \infty} \left|\frac{\frac{(n+1)^4}{2^{n+1}}}{\frac{n^4}{2^n}}\right| = \lim_{n \to \infty} \frac{(n+1)^4}{2^{n+1}}\frac{2^n}{n^4} \\ = \lim_{n \to \infty} \frac{1}{2}\frac{(n+1)^4}{n^4} \\ = \frac{1}{2}$$ which means $\sum_{i=0}^\infty \frac{n^4}{2^n}$ is convergent. And $$\lim_{n \to \infty} \left|\frac{\frac{(-1)^{k+1}(k+1)}{k+3}}{\frac{(-1)^{k}k}{k+2}}\right| = \lim_{n \to \infty}\frac{k+1}{k+3}\frac{k+2}{k} \\ = \lim_{n \to \infty} \frac{k^2+4k+3}{k^2+3k} \\ = 1$$ So the ratio test is inconclusive. $\endgroup$ – graydad Mar 28 '15 at 18:03
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    $\begingroup$ Geometric series are by far the easiest to evaluate for an exact sum. Most series are not that easy. Telescoping series are extremely easy to determine an exact value of, like $$\sum_{n=1}^\infty \sin(n)-\sin(n+1)$$ On the other hand, sums of the form $$\sum_{n=1}^\infty\frac{1}{n^s}$$ where $s$ is any fixed number (including complex numbers) can be evaluated, but not without a lot of study in number theory and complex analysis. This is not a sum you would evaluate in a calculus class. The results are bizarre and unbelievable. For example, $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ $\endgroup$ – graydad Mar 28 '15 at 18:08
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    $\begingroup$ @Rinzler happy to help! Sorry to overwhelm you, but I hope you return to read my answer when you have more free time :) As for the series you mentioned, we do not know by the ratio test that it diverges, since the limit was equal to $1$. We only know it diverges if the limit is greater than $1$. It is possible for a series to not converge and not diverge. For example, $1+(-1)+1+(-1)+1+\ldots$ will neither converge nor diverge. The series you mentioned will behave like this in the long run, so it is neither convergent nor divergent. $\endgroup$ – graydad Mar 28 '15 at 18:18

Here is a simple proof for the geometric convergence:

Suppose we have the finite sum:

$$1+r+r^2+r^3+r^4+\dots+r^k$$ with $r\neq 1$

Since this is a finite sum, it will converge to some unknown quantity: $S$

So: $$\begin{array}{lrl} S & =& 1+r+r^2+r^3+r^4+\dots+r^k\\ r\cdot S & =& ~~~~~~~ r+r^2+r^3+r^4+r^5+\dots+r^{k+1}\\ S-r\cdot S& = & 1 - r^{k+1}\\ S & = & \frac{1-r^{k+1}}{1-r}\end{array}$$

If $r=1$ then clearly $S = k+1$

Using this information for infinite series, it follows that $\sum\limits_{i=0}^\infty r^i = \lim\limits_{k\to\infty}\sum\limits_{i=0}^k r^i = \lim\limits_{k\to\infty} \frac{1-r^{k+1}}{1-r}$

In the case that $|r|>1$ this clearly diverges, and when $|r|<1$ then the $r^{k+1}$ term is insignificant, so this will converge. The case where $r=1$ followed the other form where $S = k+1$ and will clearly diverge, the case where $r=-1$ will also diverge since the sequence of partial sums will alternate between 1 and 0.

As for a general rule to find the ratio, in a geometric series, it will be of the form $\sum a \cdot r^n$. Try to algebraically manipulate the summand to be in that form. For example: $\sum 3\cdot \frac{5^n}{6^{n+1}} = \sum 3\cdot \frac{5^n}{6^n\cdot 6} = \sum \frac{3}{6}\cdot (\frac{5}{6})^n$, and your ratio will be $\frac{5}{6}$


Let $S_N=\sum_{i=0}^{N} r^i=1+r+r^2+\cdots r^N$. Note that

$$\begin{align} rS_N&=\sum_{i=0}^{N} r^{i+1}\\ &=r+r^2+r^3+\cdots+r^N+r^{N+1} \end{align}$$

So, if we form the difference $S_N-rS_N$ we find $S_N-rS_N=(1-r)S_N=1-r^{N+1}$ where upon solving for $S_N$ we find $$\begin{align} S_N&=\frac{1-r^{N+1}}{1-r}\\ \end{align}$$

If $r<1$, then $r^{N+1} \to 0$ as $N \to \infty$ and we have

$$\sum_{i=0}^{\infty} r^{i}=\frac{1}{1-r}$$.

If $r=\frac{1}{3}$, then $\sum_{i=0}^{\infty} (\frac13)^{i}=\frac{1}{1-\frac13}=\frac32$.

If the series is $4\sum_{i=0}^{\infty} (\frac13)^{i}$, then we have $4\sum_{i=0}^{\infty} (\frac13)^{i}=4\frac{1}{1-\frac13}=4\frac32 = 6$


A proof that, if the left-hand-side converges: $$a+ar+ar^2+\dotsb=\frac a{1-r}$$

Call the sum $S$. Thus, we have $S=a+ar+ar^2+\dotsb$.

$\phantom rS=a+ar+ar^2+ar^3+\dotsb$
$rS=\phantom{a+{}}ar+ar^2+ar^3+\dotsb$ (using the distributive property)

Thus: \begin{align} S&=a+rS\\ S-rS&=a\\ S(1-r)&=a\\ S&=\frac a{1-r} \end{align}


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