Finding smallest n for finite sum

**Pinkk** · May 9th 2009, 04:59 PM

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

So I set up the inequality:

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

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

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

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

**NonCommAlg** · May 9th 2009, 05:35 PM

Originally Posted by Pinkk

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

So I set up the inequality:

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

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

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

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

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

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 $n=9.$ (don't trust my calculations! do it yourself!

)

**Pinkk** · May 9th 2009, 05:40 PM

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 $\frac{2^{n+1}}{(n+1)!}$ and not $\frac{(-2)^{n+1}}{(n+1)!}$ ?

**NonCommAlg** · May 9th 2009, 06:04 PM

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 $\frac{2^{n+1}}{(n+1)!}$ and not $\frac{(-2)^{n+1}}{(n+1)!}$ ?

because $a_n$ 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 $\{a_k \}$ is

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

**Pinkk** · May 9th 2009, 06:10 PM

Ah, okay, I get it. Thanks!

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