# Thread: Calculus 2 sum and error help

1. ## Calculus 2 sum and error help

I missed a few days of my Calc 202 class and now I'm in big trouble because the final is like next week and i missed the taylor series and maclaurin and pretty much all the hardest topics i attempted to do what I could so far but I'm in a real spot now lol this problem was giving me trouble because I have no idea what it means lol I understand that the equation is a p-series with p=5 but i dont get anything else...

2. You can estimate $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^5}$ by computing $\displaystyle s_{10} = \sum_{n = 1}^{10} \frac{1}{n^5}$ (in general $\displaystyle s_k = \sum_{n=1}^k \frac{1}{n^5}$). Your estimate is going to be a little bit low since you only computed the first 10 terms of the series. $\displaystyle R_k = \int_{k}^{\infty} \frac{1}{x^5}dx$ (I think that's what this is supposed to be) gives you an upper bound on how far off $\displaystyle s_k$ is from the true value of the series. If you want to be within $\displaystyle \epsilon$, choose $\displaystyle k$ such that $\displaystyle R_k < \epsilon$.

The last part says that $\displaystyle s_k + \int_{k+1}^{\infty} \frac{1}{x^5} dx$ is guaranteed to be smaller than the true value of the series, while $\displaystyle s_k + \int_{k}^{\infty} \frac{1}{x^5} dx$ is guaranteed to be larger than the true value of the series. So you have two estimates, one which is too small and one which is too big. If you split the difference and average the two estimates, you should get something closer than either estimate. Specifically, the error will be at most half the difference between the two estimates.