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!
 A: 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 $3$.
A: 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.
