# Is this series $\sum_{n=0}^{\infty} 2^{(-1)^n - n} = 3$ or $\approx 3$

Is this series $$\sum_{n=0}^{\infty} 2^{(-1)^n - n} = 3$$ or $$\approx 3$$

Hello all. In Ross' elementary Analysis Chapter 14 there is an example of the above series. It simply says it converges. I have computed that it converges to 3. The way I did it is that I thought: $$2 + \frac{1}{4} + \frac{1}{2} + + \frac{1}{16} + + \frac{1}{8} + + \frac{1}{64} + + \frac{1}{32}....$$

can be broken up to $$2 + \sum_{n=1}^{\infty} (\frac{1}{4})^n + \frac{1}{2}\sum_{n=0}^{\infty} (\frac{1}{4})^n$$

Which is 2 + 2 geometric series that I computed to $$\frac{1}{3}$$ and $$\frac{2}{3}$$ respectively getting exactly 3. Using the formula I learned first term divided by 1 - ratio.

Now wolfram alpha says this is only approximately 3. Did I go wrong somewhere above? Or could it be that there is some rule that makes me not be able to give an exact result? I apologize if this is very basic. I have been thinking about this since yesterday.

Thanks a lot!

• Since terms are positive, you can group them as only you wish, so imo everything is fine and the result is exactly $3$. It might happen that wolfram output only approximate solution. Commented Jul 22, 2020 at 11:49
• @DominikKutek Thanks a lot! I thought it could be, but often it gave me exact answers, I use it a lot to check for solutions that aren't provided in textbook. I greatly appreciate it thanks! Commented Jul 22, 2020 at 11:50
• There is no closed form for the partial sum and you probably notice that WA returns a complex number. Commented Jul 22, 2020 at 12:06
• @ClaudeLeibovici: hi Claude. There is no problem writing a closed-form expression. You can very well split in the even and odd terms, and parity can be handled by a "parity indicator function" such as $\frac{1+(-1)^n}2$.
– user65203
Commented Jul 22, 2020 at 12:15
• @YvesDaoust. Mathematica does not return any result with Sum and with NSum there are a lot of trouble and the end result is a complex number (just as with WA). Commented Jul 22, 2020 at 12:22

Wolfram alpha is not wrong. The result is approximately 3.

You are also not wrong. The result is exactly 3.

Where you are wrong is when you use the word "only". You say:

wolfram alpha says this is only approximately 3.

This implies that you think that the number $$3$$ is not "approximately" equal to itself. In fact, it is. $$3\approx 3$$ is a true statement, so a better way of putting it would be:

wolfram alpha says only that this is approximately $$3$$

In other words, WA is not wrong, it's just not telling the entire story. It is saying that the sum is approximately $$3$$ (which means that it is either exactly $$3$$ or a number close to, but not equal to $$3$$), when in fact, it could state the logically stronger claim that the sum is exactly $$3$$.

• Thank you. Haha that makes sense, I was worried that it might be a silly question. Wolfram often gave me a exact answers so I wasn't sure here. Thank you! Commented Jul 22, 2020 at 11:53
• For the fun, try with Mathematica. Commented Jul 22, 2020 at 13:01

Alpha says Approximated sum: $$3$$, which is correct.

There is no Infinite sum: section. It turns out that Alpha does not seem to be able to compute the exact value, though the decomposition seems easy, and falls back on numerical evaluation (and if you ask more digits, you get kilometers of exact decimals).

Also consider that $$a=b\implies a\approx b$$, this is a weakening of the relation.