How many terms are needed, to estimate sum to within 0.0001? sum{n=1,infinity}(-1)^{n+1} (2/n)

[mario99]

Determine how many terms are needed to estimate the sum of the series to within [imath]0.0001[/imath].

[imath]\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n}[/imath]

By following the theorem of estimating the alternating series, my calculations are

[imath]\frac{2}{n + 1} \leq 0.0001[/imath]

[imath]n \geq 20000 - 1 = 19999[/imath]

Thus, [imath]19,999[/imath] terms are needed, but the book says [imath]20,000[/imath]! Why?

[imath]\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n} \approx 1.3863[/imath]

[imath]\sum_{n=1}^{20000}(-1)^{n+1}\frac{2}{n} \approx 1.3862[/imath]

[imath]\sum_{n=1}^{10000}(-1)^{n+1}\frac{2}{n} \approx 1.3862[/imath]

Even, [imath]n = 7000[/imath], is sufficient.

[imath]\sum_{n=1}^{7000}(-1)^{n+1}\frac{2}{n} \approx 1.3862[/imath]

Why is the answer too big while we can also get a correct estimation within [imath]0.0001[/imath] with smaller terms?

 

[topsquark]

Determine how many terms are needed to estimate the sum of the series to within [imath]0.0001[/imath].

[imath]\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n}[/imath]

By following the theorem of estimating the alternating series, my calculations are

[imath]\frac{2}{n + 1} \leq 0.0001[/imath]
...

Why is the answer too big while we can also get a correct estimation within [imath]0.0001[/imath] with smaller terms?

Remember that the error formula is an upper bound. For specific series the error may be much lower than this number.

-Dan

 

[limiTS]

You have to find the [imath]n+1[/imath] term. You already determined that [imath]n=19,999[/imath].

 

[mario99]

Thank you topsquark and limiTS for helping me.

Remember that the error formula is an upper bound. For specific series the error may be much lower than this number.

-Dan

This makes sense!

You have to find the [imath]n+1[/imath] term. You already determined that [imath]n=19,999[/imath].

I don't agree with you because all other examples whatever you get with [imath]n \geq \ [/imath] something, it is the answer unless the infinite series starts with index [imath]n = 0[/imath] (or [imath]n \neq 1[/imath]) or the answer of [imath]n[/imath] has a decimal.

For example,

[imath]\sum_{n=0}^{\infty}(-1)^{n+1}\frac{2}{n+1}[/imath], I would add 1 term to the answer.

[imath]\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{n^4}[/imath], Here [imath]n \geq \sqrt[4]{10^{10}} - 1 \approx 315.2[/imath], I would round up so that, [imath]n \geq 316[/imath]

Therefore, do you think that it is just a fancy way to write [imath]n = 20,000[/imath], rather than [imath]n = 19,999[/imath] when both answers are valid and the former is much clearer for the eyes to see?

 

[limiTS]

The book's official answer of 20,000 terms is correct, not 19,999, even if fewer terms get you within 0.0001.

 

[Dr.Peterson]

Thus, 19,999 terms are needed, but the book says 20,000! Why?

Can you show us the exact wording of the book's question and answer?

The only explanation I can think of for their answer would be if "within 0.0001" were taken to mean that the last term has to be less than 0.0001.

Maybe you can show us a worked example of the same type, to see if there is anything different in the way they perceive the theorem applying.

 

[limiTS]

I tested the OP's claim about the 7,000th term being within 0.0001 of the infinite sum, which happens to converge to [imath]\ln 4[/imath].

The 7,000th partial sum is approx. 1.386151514, whereas [imath]\ln 4\approx1.386294361[/imath], so the error is approx. 0.000142847. Unfortunately, this is still [imath]>0.0001[/imath]

The 20,000th partial sum is approx. 1.386244362, and the error is approx. 0.00004999875, which is [imath]<0.0001[/imath]. Like topsquark said, the error formula is only an upper bound, but fewer terms could be enough to get within 0.0001. It looks like the 10,000th partial sum is finally within 0.0001. Sorry, 9,999.

 

[mario99]

The book's official answer of 20,000 terms is correct, not 19,999, even if fewer terms get you within 0.0001.

the book chose [imath]20,000[/imath] terms, not necessarily [imath]19,999[/imath] terms are wrong.

I will show you other ways the book used to solve similar problems. We will compare the new examples with our original post.

Can you show us the exact wording of the book's question and answer?

The only explanation I can think of for their answer would be if "within 0.0001" were taken to mean that the last term has to be less than 0.0001.

Maybe you can show us a worked example of the same type, to see if there is anything different in the way they perceive the theorem applying.

Determine how many terms are needed to estimate the sum of the series to within [imath]0.0001[/imath].

This is the exact wording.

OK. I will show you two examples as well as our original post. The two examples will be within [imath]0.01[/imath].

[imath]\sum_{n=1}^{\infty}(-1)^{n+1}\frac{4}{n^3}[/imath]

[imath]|S - S_n| \leq a_{n+1} = \frac{4}{(n+1)^3} \leq 0.01[/imath]

If [imath]n = 7[/imath], [imath]a_8 = \frac{4}{8^3} \approx 0.007813\leq 0.01[/imath]

And the book says, [imath]7[/imath] terms are needed to estimate the sum within [imath]0.01[/imath]

[imath]\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n^3}[/imath]

[imath]|S - S_n| \leq a_{n+1} = \frac{2}{(n+1)^3} \leq 0.01[/imath]

If [imath]n = 5[/imath], [imath]a_6 = \frac{2}{6^3} \approx 0.009259 \leq 0.01[/imath]

And the book says, [imath]5[/imath] terms are needed to estimate the sum within [imath]0.01[/imath]

We will follow the same process with the original post.

[imath]\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n}[/imath]

[imath]|S - S_n| \leq a_{n+1} = \frac{2}{n+1} \leq 0.0001[/imath]

If [imath]n = 19,999[/imath], [imath]a_{20,000} = \frac{2}{20,000} = 0.0001 \leq 0.0001[/imath]

The book must have said, [imath]19,999[/imath] terms are needed to estimate the sum within [imath]0.0001[/imath]

 

[limiTS]

You're right, mario99. The answer is 19,999 terms. I misremembered the error formula.

 

[Dr.Peterson]

I tested the OP's claim about the 7,000th term being within 0.0001 of the infinite sum, which happens to converge to [imath]\ln 4[/imath].

The 7,000th partial sum is approx. 1.386151514, whereas [imath]\ln 4\approx1.386294361[/imath], so the error is approx. 0.000142847. Unfortunately, this is still [imath]>0.0001[/imath]

The 20,000th partial sum is approx. 1.386244362, and the error is approx. 0.00004999875, which is [imath]<0.0001[/imath]. Like topsquark said, the error formula is only an upper bound, but fewer terms could be enough to get within 0.0001. It looks like the 10,000th partial sum is finally within 0.0001. Sorry, 9,999.

To make sure things are clear, all this is irrelevant to the question as actually intended, though it is stated in a somewhat misleading way. (I only know what was intended by knowing typical problems of this sort, and the vicinity of their answer.)

They asked how many terms are needed, but as we've seen, far fewer than 20,000 terms are really necessary to make the sum be closer than 0.0001.

On the other hand, they didn't ask about making the actual sum that close, but about estimating it that accurately. And they imply that they are talking about using the particular theorem they have just taught to decide how many terms to use; there may well be a different theorem or method.

So in terms of the question itself (as opposed to understanding how series actually work), it makes no difference what number of terms actually produces such an accurate sum. They intend to ask, how many terms does the theorem require you to use in your estimate.

Turning now to the OP:

Determine how many terms are needed to estimate the sum of the series to within 0.0001.

This is the exact wording.

That's what I expected. It didn't sound like you paraphrased the problem, but I wanted to be sure. What I said above does apply.

Thanks for this information.

OK. I will show you two examples as well as our original post. The two examples will be within [imath]0.01[/imath].

[imath]\sum_{n=1}^{\infty}(-1)^{n+1}\frac{4}{n^3}[/imath]

[imath]|S - S_n| \leq a_{n+1} = \frac{4}{(n+1)^3} \leq 0.01[/imath]

If [imath]n = 7[/imath], [imath]a_8 = \frac{4}{8^3} \approx 0.007813\leq 0.01[/imath]

And the book says, [imath]7[/imath] terms are needed to estimate the sum within [imath]0.01[/imath]

It's interesting that they don't show how they actually found this number; they just show the check. (They could have done the same check with a number greater than 7, and it would work; they haven't shown that 7 is the [minimum] number needed (according to the theorem, much less by actual summation).

Presumably, they solved the inequality [imath]\frac{4}{(n+1)^3} \leq 0.01[/imath]. This becomes [imath]4\leq0.01(n+1)^3[/imath], so that [imath]400\leq(n+1)^3[/imath], and then [imath]n+1\geq400^{1/3}>7[/imath]. Therefore the smallest value of n is 7.

In any case, yes, you did just what they did.

My suspicion is that they (explicitly or implicitly) used a "less than", and solved [imath]\frac{2}{n+1} < 0.0001[/imath], for which the smallest integer solution is 20,000, because 19,999 makes it exactly equal. One way or another, this seems to have thrown them off.

But it didn't trick you.

 

[mario99]

You're right, mario99. The answer is 19,999 terms. I misremembered the error formula.

Thank you limiTS.

It's interesting that they don't show how they actually found this number

In fact, they did show how as I did in the first post to find [imath]n \geq 19,999[/imath].

They used both methods alternatively. If the infinite series is simple so that you can isolate [imath]n[/imath], you are OK (have the freedom) to use any method. On the other hand, the trial and error method will be so effective for series such as

[imath]\sum_{n=1}^{\infty}(-1)^n\frac{2^n}{n!}[/imath]

where the factorial with or without exponential is present which makes it very difficult to isolate [imath]n[/imath].

My suspicion is that they (explicitly or implicitly) used a "less than", and solved [imath]\frac{2}{n+1} < 0.0001[/imath], for which the smallest integer solution is 20,000, because 19,999 makes it exactly equal. One way or another, this seems to have thrown them off.

But it didn't trick you.

My suspicion is the same. I think that they have solved tons of problems and they always found the term [imath]a_{n+1} <[/imath] the error, which makes the brain unconsious to accept [imath]a_{n+1} = [/imath] the error.


Thank you Dr.Peterson for the help and I also thank topsquark and limiTS for igniting the spark.