## wcaprar 3 years ago Calc II Alternating Remainder Estimate Help Needed

1. wcaprar

Determine the smallest number of terms required to approximate the sum of the series with an error less than 0.0001. $\sum_{n}^{\infty}(-7^k)/(6^kk!)$

2. wcaprar

I understand that you would set the sum equal to .0001 and solve for k. Right?

3. wcaprar

I can simplify the equation by saying (1/k!)x(7/6)^k right?

4. bmp

If I understood the problem correctly, there are a couple of ways to do it. Remember that the remainder estimate for alternating series is given by:$|R_n| \le b_{n+1}$So yeah, you can simplify it and put in the form:$0,0001 \le \frac{7^{k+1}}{6^{k+1}(k+1)!} = \frac{1}{(k+1)!}*(\frac{7}{6})^{k+1}$Or if you have a calculator, you can start expanding some terms until one of them is < 0.0001

5. wcaprar

I did that and I got to the seventh term, it was the right answer! Thanks! But on the test I won't be able to use a calculator, can this problem be solved some way w/o one?

6. bmp

If they ask for the smallest number, I guess you have to 1) manually compute the values or 2) solve an equation of the form above. If they do not, just pick a reasonable number and compute it, like n = 10. Very likely the error will be really small and it would suffice the boundary for |Rn|

7. bmp

And by the way, doing a series test without a calculator sucks a lot, huh

8. eliassaab

I agree with @bmp and add that you must fire your teacher.

9. wcaprar

Im not sure if we will be given this same type of problem but we did go over a problem like this in class that was WAY easier that we solved with very little effort

10. eliassaab

Why should one compute 7^k for k=1 to 7 and similar problems by hand. Those antiquated teachers should be trained to undersatand that the world has changed and they should follow the flow.