# Finding formula for the nth partial sum

A few days ago, I asked for some clarification about pattern recognition and the n-th partial sum for infinite series. Although the explanation given was top-notch (thanks again), I'm still having difficulty with the homework. The one I'm asking about tonight is the following sequence: $1 - 2 + 4 - 8 + ... + (-1)^{n-1}2^{n-1} + ...$

In my efforts to solve this problem, this is what I've gotten thus far: $$\begin{array}{lcc} \textrm{Parial Sum} & \textrm{Value} & \textrm{Suggested Expression} \\ s_1 = 1 & 1 & ?? \\ s_2 = 1 - 2 & -1 & ??\\ s_3 = 1 - 2 +4 & 3 & ??\\ s_4 = 1 -2 +4 -8 & -5 & ??\\ \end{array}$$

As you can see from the question marks where suggested expressions might be, I'm struggling to find the pattern. What I do know is the formula to compute $a_n$, but I haven't discerned the pattern for the n-th partial sum. Because this is a power of 2 series, I see that the magnitude between the values of each sum is exactly the power of 2 for the next n-1. That is, the distance between 1 and -1 is $2^1$ and the distance between -1 and 3 is $2^2$. I think that within this is the key to figuring this out. Nevertheless, the solution eludes me and I need a hint.

One of my attempts was $(-1)^{n-1}*2(\frac{1}{2^{n-1}})$ which worked for the first two partial sums but then fell apart miserably. While typing this up, I just made the further discovery that starting with $s_2$ each partial sum is equal to $2^n - m$ where m is the same number twice. I know that probably doesn't make sense but $s_2, s_3$ are both equal to $2^n - 5$ and $s_4, s_5$ are both $2^n - 21$. The next two are $2^n - 85$. That can't be coincidence.

Please, help me see what I'm missing or help me to understand how I should set this up to find the pattern for the n-th partial sum.

Thanks, Andy

-
YOu can show that if $r\neq 1$ then $$1+r+r^2+\cdots+r^n = \frac{r^{n+1}-1}{r-1}.$$ –  tetori Apr 16 '13 at 3:02
Adding more lines to the table might make it easier. A spreadsheet and copy down works very well. When you got higher you could see more easily that each one is $-2$ times the last plus $(-1)^{n+1}$ –  Ross Millikan Apr 16 '13 at 3:15

Later data often show patterns better than early data, so extend your table of partial sums a bit:

$$\begin{array}{rcc} n:&1&2&3&4&5&6&7&8&9\\ s_n:&1&-1&3&-5&11&-21&43&-85&171 \end{array}$$

Ignoring the signs, it appears that the numbers in the bottom line are approximately doubling each time. Moreover, still ignoring signs, adjacent partial sums add up to a power of $2$: $|s_1|+|s_2|=2^1$, $|s_2|+|s_3|=2^2$, $|s_3|+|s_4|=2^3$, and apparently in general $|s_n|+|s_{n+1}|=2^n$. (If you go back to the definition of the partial sums, you’ll see why this happens.)

If $|s_n|+|s_{n+1}|=2^n$ and $|s_{n+1}|\approx 2|s_n|$, then $3|s_n|\approx 2^n$; this suggests that we should compare $3|s_n|$ with $2^n$:

$$\begin{array}{rcc} n:&1&2&3&4&5&6&7&8&9\\ s_n:&1&-1&3&-5&11&-21&43&-85&171\\ 3|s_n|:&3&3&9&15&33&63&129&255&513\\ 2^n:&2&4&8&16&32&64&128&256&512 \end{array}$$

That pattern’s pretty clear: apparently $3|s_n|=2^n+1$ if $n$ is odd, and $3|s_n|=2^n-1$ if $n$ is even. Those cases can be combined as $3|s_n|=2^n+(-1)^{n+1}$, since $(-1)^{n+1}$ is $1$ when $n$ is odd and $-1$ when $n$ is even. And the algebraic sign of $s_n$ appears to be that of $(-1)^{n+1}$, so if these patterns are real,

\begin{align*} s_n&=\frac{(-1)^{n+1}}3\left(2^n+(-1)^{n+1}\right)\\ &=\frac{(-1)^{n+1}2^n}3+\frac{(-1)^{2n+2}}3\\ &=\frac{(-1)^{n+1}2^n+1}3\\ &=\frac{1-(-2)^n}3\;. \end{align*}

This result can then be proved by mathematical induction, but I suspect that you’re not expected to go that far.

If you’ve already learned the summation formula for finite geometric series, you can apply it to get $s_n$ without looking at any patterns at all, and it’s something that you should learn as soon as possible if you don’t already know it. However, skill at pattern-recognition is useful anyway, so I thought that it might be useful to see how the problem can be attacked in that way as well.

-
Once again, I'm grateful to you, @brianmscott, for this answer. I really appreciate the detail. I'm marking this answer as the answer only because it does do what this homework problem is addressing from the section. The formula you linked to, and the other respondent used, is in the section that my prof is attacking this Thursday. Definitely, a handy tool to keep from this trouble. I hope that a problem like this isn't on the final. I'm struggling too much for any confidence one like this. Thanks again. –  Andrew Falanga Apr 16 '13 at 19:21
@Andrew: You’re welcome. I can’t spead for your professor, but if I were designing the final, I’d be more interested in testing the material for which this is preparation than in testing this. –  Brian M. Scott Apr 16 '13 at 23:23
We have $$1+r+r^2 + \cdots + r^{n-1} = \dfrac{1-r^n}{1-r}, \,\,\,\,\, \forall r \neq 1 \tag{\star}$$ In your case, $r=-2$.
Hence, we get that the partial sum is$$s_n = \dfrac{1-(-2)^n}3$$
You may be interested in this question for more details on why $(\star)$ is true and what happens to its infinite sum, when $\vert r \vert < 1$.
@julien :) ${}{}{}$ –  user17762 Apr 16 '13 at 6:16