Alternating series convergence and error bound for ∑(-1)^k k^10/10^k

[Sheneron]

Homework Statement

Show that the series converges. Then compute an estimate of the limit that is guaranteed to be in error by no more than 0.005

$$\sum_{k=5}^{\infty} (-1)^k \frac{k^{10}}{10^k}$$

The Attempt at a Solution

This is obviously an alternating series and I know that

$$C_{k} = \frac{k^{10}}{10^k}$$

and I know that Cn+1 is greater than the absolute value of S-Sn. So I can set up to be something like

$$\frac{(n+1)^{10}}{10^{n+1}} <= 0.005$$

the part I can't figure out is how to solve that for n. Is there a way to simplify that fraction? How would I solve this for n? Thanks

Oh and I also couldn't figure out how to exactly show that the limit as k-> infinity of Ck goes to 0 without taking the derivative 10 times. So the whole problem I am having is with the fraction.

 

[Dick]

If you want to show the limit goes to zero, take the log and try showing that the limit of that goes to -infinity. Once you done that also find the value of x where x^10/10^x is a maximum by maximizing the log. Then you know it's decreasing after that value. No, I don't think you can really 'solve for n'. But once you know where the series is decreasing, just find an n so the term is less that 0.005.

 

[Sheneron]

Thanks, that makes sense.