I'm learning now about sequences and series: patterns in short. This is part of my Calc II class. I'm finding I'm having difficulty in detecting all of the patterns that my text book is asking me to solve. My question at this point isn't directly about a homework problem (yet anyway), but instead help in understanding why certain statements are made in the example.

So, the example from the book:

$$ 1 + \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\\ \begin{array}{lcc} Partial Sum & Value & Sugg. Expression \\ s_1 = 1 & 1 & 2 - 1 \\ s_2 = 1 + \frac{1}{2} & \frac{3}{2} & 2 - \frac{1}{2} \\ s_3 = 1 + \frac{1}{2} + \frac{1}{4} & \frac{7}{4} & 2 - \frac{1}{4} \\ & ... & \\ s_n = 1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{n-1}} & \frac{2^n - 1}{2^{n-1}} & 2 - \frac{1}{2^{n-1}} \end{array} $$ Why is that for $s_1$ they say, "Suggested Expression" is 2-1? 4 - 3 also yields 1. Granted, the suggested values in the textbook are much simpler to work with. However, I'd like to know what wisdom leads the authors to say that 2-1 is the suggested expression instead of some other expression also yielding 1.

It is also interesting to me that this process is necessary when this sequence is quite easily seen as $\sum_{n=1}^{\infty}\frac{1}{2^{n-1}}$. This section is all about convergence and divergence. Am I learning these extra steps because using the rule I've just outlined doesn't show what it converges to? Also, in typing this question, I think I've just discovered something the textbook was saying: a series isn't convergent unless the limit of its terms is 0. That is: $\lim_{n\to\infty}\frac{1}{2^{n-1}} = 0$. It's amazing what one finds when looking for something else.

Thanks, Andy

  • $\begingroup$ Near the end of the last paragraph, that should be "a series isn't convergent unless the limit of its terms is $0$. That is: $\lim_{n\to\infty}\frac1{2^{n-1}}=0$". The limit you wrote is meaningless because the $n$ you're trying to take to $\infty$ is a dummy summation variable that can't appear outside the summation. $\endgroup$ – joriki Apr 11 '13 at 5:00
  • $\begingroup$ Thank you @joriki I've fixed it with an edit. $\endgroup$ – Andrew Falanga Apr 12 '13 at 4:21
  • $\begingroup$ You didn't; you fixed one half of it. $\endgroup$ – joriki Apr 12 '13 at 8:55
  • 1
    $\begingroup$ @joriki ok, I think I fixed the other half $\endgroup$ – Andrew Falanga Apr 15 '13 at 20:07

The suggested expressions weren’t found one at a time: they’re synthesized from the whole pattern. We don’t seriously consider $4-3$ for the first one, for instance, because it doesn’t fit nicely with anything else. I’d describe the thought process this way. First we calculate the first few partial sums:

$$\begin{array}{r|cc} n&1&2&3&4&5&6\\ \hline s_n&1&\frac32&\frac74&\frac{15}8&\frac{31}{16}&\frac{63}{32} \end{array}$$

At this point I can pursue either of two lines of thought.

  1. The partial sums seem to be getting very close to $2$. Perhaps they’re doing so in some regular, easily identifiable fashion? Let’s add another line to the table: $$\begin{array}{r|cc} n&1&2&3&4&5&6\\ \hline s_n&1&\frac32&\frac74&\frac{15}8&\frac{31}{16}&\frac{63}{32}\\ \hline 2-s_n&1&\frac12&\frac14&\frac18&\frac1{16}&\frac1{32} \end{array}$$ Now that was very informative: the denominators of the new entries are instantly recognizable as powers of $2$, specifically $2^{n-1}$, and it looks very much as if $2-s_n=\frac1{2^{n-1}}$, or $$s_n=2-\frac1{2^{n-1}}\;.$$ This is the line of thought that leads to the suggested expressions in the example.

  2. The denominators of $s_n$ are instantly recognizable as powers of $2$, specifically $2^{n-1}$, and the numerators seem to be one less than the next higher power of $2$, or $2^n-1$. It looks very much as if $$s_n=\frac{2^n-1}{2^{n-1}}\;.$$

A little algebra of course shows that the conjectures are the same: $2-\dfrac1{2^{n-1}}=\dfrac{2^n-1}{2^{n-1}}$.

Without seeing the example in full I can’t be sure, but I suspect that the suggested expressions are there because they make it immediately evident that


since clearly $\lim\limits_{n\to\infty}\dfrac1{2^{n-1}}=0$. Essentially the same idea is at the heart of the proof that

$$\sum_{n\ge 0}x^n=\frac1{1-x}$$

if $|x|<1$; it that argument hasn’t yet appeared in your text, this example may be part of the preparation.

Finally, note the correction of your final remark that joriki made in the comments.

  • $\begingroup$ @brianmscott Thank you very much. This helps a great deal. Especially, the detailed description in "thought process" 1. After spending a few more hours on these things, I now see that what I was confusing before was that I'm trying to find the rule that allows me to calculate the n-th partial sum rather than calculate the n-th term. It would seem that there isn't a "hard & fast" rule for discovery of the patterns. One needs to spend time with these to learn how to decipher them. Thanks again. $\endgroup$ – Andrew Falanga Apr 12 '13 at 4:29
  • $\begingroup$ @Andrew: You’re welcome. $\endgroup$ – Brian M. Scott Apr 12 '13 at 4:50

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