[SOLVED] Find the least number of n

**mark1950** · August 28th 2009, 07:27 AM

Given the geometric series, $1,-\frac{1}{2},\frac{1}{4}...$ find the least number of n so that the difference between the Sum to infinity and the Sum to n is less than $10^-5$

This is my take:
$r = -\frac{1}{2}$

$a = 2$

Substituting r and a into the formula of Sum to infinity and Sum to n,

$\mid\frac{2}{1+\frac{1}{2}} - \frac{2[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}\mid< 10^{-5}$

$\mid\frac{4}{3} - \frac{2[1-(-\frac{1}{2})^n]}{\frac{3}{2}}\mid < 10^{-5}$

$\mid\frac{4}{3} - \frac{4}{3}[1-(-\frac{1}{2})^n]\mid < 10^{-5}$

$\mid\frac{4}{3}[1-[1-(-\frac{1}{2})^n]]\mid < 10^{-5}$

$[\frac{4}{3}(-\frac{1}{2})^n]^2 < [10^{-5}]^2$

$[\frac{16}{9}(-\frac{1}{2})^{2n}] < [10^{-10}]$

$[(-\frac{1}{2})]^{2n} < [5.625 \times 10^{-11}]$

$\lg [(-\frac{1}{2})]^{2n} < \lg[5.625 \times 10^{-11}]$

Then, I decided to use logarithms but when I typed it onto the calculator, I found out that it was not possible (maths error) because $\lg [(-\frac{1}{2})]$ = undefined. Anyone know where I had gone wrong? Thanks.

**RobLikesBrunch** · August 28th 2009, 08:29 AM

Note: If you can't find a suitable explanation for the whole alternating series deal, just tell me and I'll go through it. I'm just being incredibly lazy right now.

I think you're taking the wrong approach. (nevermind about this)

Let's take the negative out of r and introduce it seperately into the formula (I also changed a=1):

$r = \frac{1}{2}$

$\displaystyle\sum_{n=0}^{\infty} (-1)^{n} (\frac{1}{2})^{n} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...$

This is clearly an alternating series, and it satifies the Alternating Series Test for convergence, and therefore, we can utilize the Alternating Series Estimation Theorem. (Google these things if you aren't familiar with them).

...but now that I think about it, the series' values are fairly "strong" for a while, so you have to go through quite a lot of them.

Anyway, I went through them, and by the time you hit $\frac{1}{16384} = 0.000061$ (I don't know how to make the approximate sign in LaTeX...and here I'm talking only about b-sub-n part..where n=14[again, don't know the LaTeX for it]) you're at a point less than $10^{-5}$ . So we have to go up to the point n = 13 as at this point, b-sub-n equals $\frac{1}{8192}$ ...which still modifies the 4th decimal place, but that's okay because the next value in the series does not.

But I don't think this approach is best....so nevermind...if you're curious how I found it out, just look up the theorems (if you didn't know them beforehand)...and I don't understand your approach at all...so I automatically think it's vastly more elegant than mine.

**Opalg** · August 28th 2009, 08:39 AM

Originally Posted by mark1950

Given the geometric series, $1,-\frac{1}{2},\frac{1}{4}...$ find the least number of n so that the difference between the Sum to infinity and the Sum to n is less than $10^{-5}$

This is my take:
$r = -\frac{1}{2}$

$a = 2$ Should be a = 1, but that doesn't really affect the calculation much.

Substituting r and a into the formula of Sum to infinity and Sum to n,

$\left|\frac{2}{1+\frac{1}{2}} - \frac{2[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}\right|< 10^{-5}$

$\left|\frac{4}{3} - \frac{2[1-(-\frac{1}{2})^n]}{\frac{3}{2}}\right| < 10^{-5}$

$\left|\frac{4}{3} - \frac{4}{3}\Bigl[1-\Bigl(-\frac{1}{2}\Bigr)^n\Bigr]\right| < 10^{-5}$

$\left|\frac{4}{3}\Bigl[1-\Bigl[1-\Bigl(-\frac{1}{2}\Bigr)^n\Bigr]\Bigr]\right| < 10^{-5}$

This is the stage where it starts to go astray. Instead of squaring both sides, write that last line as $\left|\frac{4}{3}\Bigl(-\frac{1}{2}\Bigr)^n\right| < 10^{-5}$ . To take the absolute value of the left side, all you have to do is to drop the minus sign, so that it becomes $\frac{4}{3}\Bigl(\frac{1}{2}\Bigr)^n < 10^{-5}$ .

You'll still need to be careful when taking logarithms, because log(1/2) is negative. So it might be better to take the reciprocal of both sides and write the inequality as $\tfrac34\cdot2^n > 10^5$ . That should be something that the calculator can cope with.

**RobLikesBrunch** · August 28th 2009, 08:58 AM

Originally Posted by Opalg

This is the stage where it starts to go astray. Instead of squaring both sides, write that last line as $\left|\frac{4}{3}\Bigl(-\frac{1}{2}\Bigr)^n\right| < 10^{-5}$ . To take the absolute value of the left side, all you have to do is to drop the minus sign, so that it becomes $\frac{4}{3}\Bigl(\frac{1}{2}\Bigr)^n < 10^{-5}$ .

You'll still need to be careful when taking logarithms, because log(1/2) is negative. So it might be better to take the reciprocal of both sides and write the inequality as $\tfrac34\cdot2^n > 10^5$ . That should be something that the calculator can cope with.

Can you tell me if anything that I did was even remotely correct? Also, what's the formula called that the OP is talking about?

**mark1950** · August 28th 2009, 09:10 AM

@Opalg

Yea. a = 2 is correct. My sequence was wrongly typed because I saw the wrong question. Sorry about that.

Btw, what's the logic behind dropping the minus sign in a bracket? I don't understand that part. I know there's the absolute sign there but I still don't get it. Care to give simple examples? Thanks. Also, why it is not correct to square both sides? I thought squaring both sides should be applicable to all kinds of equalities involving the absolute sign? Thanks.

@ RobLikesBrunch

There are two formulas used. Sum to infinity = $\frac{a}{1 - r}$ and Sum to the nth term = $\frac{a(1-r^n)}{1-r}$ for $\mid r \mid < 1$ or $\frac{a(r^n-1)}{r-1}$ for $\mid r \mid > 1$ where n = number of terms, a = first term, and r = common ratio. These formulas are only applicable for geometric progressions. There is another progression called the arithmetic progression which uses the same symbols except for the r which is replaced by d and its formulas are not the same as the geometric one.

**Opalg** · August 28th 2009, 10:30 AM

Originally Posted by RobLikesBrunch

Can you tell me if anything that I did was even remotely correct?

Your approach is fine, nothing wrong with it. It is based on the fact that the series is an alternating series. My comment followed the OP's method, using the fact that it is a geometric series. Both approaches are equally good, and should give the same result.

My only quibble with your solution is that 0.000061 is not less than $10^{-5}$ . In fact it is $6.1\times10^{-5}$ . So you need an extra decimal place to get the correct solution.

Originally Posted by mark1950

@Opalg

Btw, what's the logic behind dropping the minus sign in a bracket? I don't understand that part. I know there's the absolute sign there but I still don't get it. Care to give simple examples? Thanks. Also, why it is not correct to square both sides? I thought squaring both sides should be applicable to all kinds of equalities involving the absolute sign? Thanks.

The absolute value of a negative number, by definition, means the positive number of the same magnitude. So for example |–7| = 7. Also $|-\tfrac12| = \tfrac12$ , $|\bigl(-\tfrac12\bigr)^5| = \bigl(\tfrac12\bigr)^5$ , and $|\bigl(-\tfrac12\bigr)^n| = \bigl(\tfrac12\bigr)^n$ .

Squaring both sides is one way to get rid of the absolute value signs. There's nothing incorrect about doing that, but it's not always the simplest method.

**mark1950** · August 28th 2009, 08:02 PM

Hm...if I were to square both sides, what would happen? I tried squaring both sides but I can't seem to get rid of the negative sign. Thanks.

**Opalg** · August 28th 2009, 11:25 PM

Originally Posted by mark1950

Hm...if I were to square both sides, what would happen? I tried squaring both sides but I can't seem to get rid of the negative sign. Thanks.

If you square both sides of the inequality $\left|\frac{4}{3}\Bigl(-\frac{1}{2}\Bigr)^n\right| < 10^{-5}$ then it becomes $\frac{16}9\Bigl(-\frac{1}{2}\Bigr)^{2n} < 10^{-10}$ . The negative sign is still there. But notice that the fraction –1/2 is raised to an even power (2n). If you raise a negative number to an even power, it gives the same answer as when you raise the corresponding positive number to the same power. So $\Bigl(-\frac{1}{2}\Bigr)^{2n} = \Bigl(\frac{1}{2}\Bigr)^{2n}$ .

**mark1950** · August 29th 2009, 01:50 AM

Thanks! You guys have helped me a lot. Thread solved.

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