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
Best Answer
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.