Thread: Finding smallest n for finite sum

1. Finding smallest n for finite sum

Find smallest n so that the finite sum $\displaystyle \sum_{k=1}^{n}\frac{(-2)^{k}}{k!}$ is certain to differ from $\displaystyle \sum_{k=1}^{\infty}\frac{(-2)^{k}}{k!}$ by less than .001.

So I set up the inequality:

$\displaystyle a_{n+1} \le .001$

$\displaystyle \frac{(-2)^{n+1}}{(n+1)n!} \le \frac{1}{100}$

$\displaystyle (n+1)n! \ge -200(-2)^{n}$

And that is where I am stuck. How do I solve for $\displaystyle n$?

2. Originally Posted by Pinkk
Find smallest n so that the finite sum $\displaystyle \sum_{k=1}^{n}\frac{(-2)^{k}}{k!}$ is certain to differ from $\displaystyle \sum_{k=1}^{\infty}\frac{(-2)^{k}}{k!}$ by less than .001.

So I set up the inequality:

$\displaystyle a_{n+1} \le .001$

$\displaystyle \frac{(-2)^{n+1}}{(n+1)n!} \le \frac{1}{100}$

$\displaystyle (n+1)n! \ge -200(-2)^{n}$

And that is where I am stuck. How do I solve for $\displaystyle n$?
note that $\displaystyle a_{n+1}=\frac{2^{n+1}}{(n+1)!}$ not $\displaystyle \frac{(-2)^{n+1}}{(n+1)!}.$ the inequality $\displaystyle \frac{2^{n+1}}{(n+1)!}< .001$ cannot be solved for $\displaystyle n$ algebraically. however, it is possible to make it a little simpler by finding a lower bound for $\displaystyle \frac{2^{n+1}}{(n+1)!},$

which is not worth the effort anyway. the easiest way, and you're expected to do it this way, to use your calculator. i think you'll get $\displaystyle n=9.$ (don't trust my calculations! do it yourself! )

3. We're not allowed to use calculators for any of our exams, so my professor must have accidentally entered this problem. And why is it $\displaystyle \frac{2^{n+1}}{(n+1)!}$ and not $\displaystyle \frac{(-2)^{n+1}}{(n+1)!}$?

4. Originally Posted by Pinkk

We're not allowed to use calculators for any of our exams, so my professor must have accidentally entered this problem.
you won't have such a problem in your exam then! haha

And why is it $\displaystyle \frac{2^{n+1}}{(n+1)!}$ and not $\displaystyle \frac{(-2)^{n+1}}{(n+1)!}$?
because $\displaystyle a_n$ is always the absolute value of the n-th term of your alternating series. the theorem that they taught you, and you should take a look at it again, is that if the sequence $\displaystyle \{a_k \}$ is

positive, decreasing and $\displaystyle \lim_{k\to\infty} a_k=0,$ then for any $\displaystyle n: \ \left|\sum_{k=1}^{\infty} (-1)^k a_k - \sum_{k=1}^n (-1)^k a_k \right| \leq a_{n+1}.$

5. Ah, okay, I get it. Thanks!