Remainder Estimate for Integral Test

[Pull and Twist]

I'm working on the following problem and I have made it this far... am I on the correct path or am I doing this incorrectly?? I find series extremely confusing. Also... how do I find the error involved in the improved approximation?

This is the series I am working with: \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{{n}^{1.01}}\)

View attachment 4249

 

[chisigma]

I'm working on the following problem and I have made it this far... am I on the correct path or am I doing this incorrectly?? I find series extremely confusing. Also... how do I find the error involved in the improved approximation?

This is the series I am working with: \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{{n}^{1.01}}\)

Take into account the Laurent series expansion discovered by the Dutch mathematician Thomas Joannes Stieltjes in the XIX century...

$\displaystyle \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} = \frac{1}{s-1} + \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} (s - 1)^{n},\ |s|>1\ (1) $

It is clear that if s is close to 1 the first term will be dominant, so that $\displaystyle \zeta (1.01) \sim 100$ ... if You want to take into account the term of order 0, the coefficient $\gamma_{0}$ in (1) is known as the Euler's constant $\gamma = .57721566...$ so that is $\displaystyle \zeta (1.01) \sim 100. 57721566... $

... it is clear that also adding other terms the result will not deviate too much...

Kind regards

$\chi$ $\sigma$

 

[Pull and Twist]

Take into account the Laurent series expansion discovered by the Dutch mathematician Thomas Joannes Stieltjes in the XIX century...

$\displaystyle \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} = \frac{1}{s-1} + \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} (s - 1)^{n},\ |s|>1\ (1) $

It is clear that if s is close to 1 the first term will be dominant, so that $\displaystyle \zeta (1.01) \sim 100$ ... if You want to take into account the term of order 0, the coefficient $\gamma_{0}$ in (1) is known as the Euler's constant $\gamma = .57721566...$ so that is $\displaystyle \zeta (1.01) \sim 100. 57721566... $

... it is clear that also adding other terms the result will not deviate too much...

Kind regards

$\chi$ $\sigma$

I have never learned about the Laurent series expansion... I'm supposed to be using the Remainder Estimate for Integral Test... I am assuming that I am not using it correctly since I am not getting the same answer.

 

[Euge]

I'm working on the following problem and I have made it this far... am I on the correct path or am I doing this incorrectly?? I find series extremely confusing. Also... how do I find the error involved in the improved approximation?

This is the series I am working with: \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{{n}^{1.01}}\)

Hi PullandTwist,

There are many errors in your work. While you can approximate $s$ using $\int_1^\infty dx/x^{1.01}$, it is not the case that $s = \int_1^\infty dx/x^{1.01}$, which is what your work shows on the left side of the paper. The main point though is that it has nothing to do with the problem; you should remove it. In the problem, you must approximate $s$ with $s_1$ (which you did -- you have $s_1 = 1$) and then use the estimates for $R_1$ to give the error of approximation (it should be fine to give the order of magnitude of $R_1$). In your analysis, you gave an approximation of $s$ of about $75.827$. This should be a red flag, because the checkpoint was $s \approx 100.65$. Knowing that $\int_{n+1}^\infty x^{-1.01}\, dx \le R_n \le \int_n^\infty x^{-1.01}\, dx$, after computing the integrals we obtain $100 (n+1)^{-0.01} \le R_n \le 100n^{-0.01}$. Substituting $n = 1$, we get $100(0.5)^{-0.01} \le R_1 \le 100$. What does this tell you about the order of magnitude of $R_1$?

 

[chisigma]

I have never learned about the Laurent series expansion... I'm supposed to be using the Remainder Estimate for Integral Test... I am assuming that I am not using it correctly since I am not getting the same answer.

... what You can obtain from integral test is...

$\displaystyle |R_{n}| \le \int_{n}^{n+1} x^{-1.01}\ d x = \frac {100}{99}\ \{ (n+1)^{\frac{99}{100}} -n^{\frac{99}{100}}\}\ (1) $

... so that if $s_{0}=100$ You have $|R_{1}|\le .996...,\ |R_{2}| \le .9909..., |R_{3}|\le .9875...$ and so on...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

I'm working on the following problem and I have made it this far... am I on the correct path or am I doing this incorrectly?? I find series extremely confusing. Also... how do I find the error involved in the improved approximation?

This is the series I am working with: \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{{n}^{1.01}}\)

You may find these notes helpful – see the section headed Integral Test.

The setting is as follows. Suppose that $f(x)$ is a positive, decreasing function defined for $x>0$ and let $a_n = f(n).$ Let \(\displaystyle s = \sum_{j=1}^\infty a_j\) and let \(\displaystyle s_n = \sum_{j=1}^n a_j.\) So $s$ denotes the sum of the whole series, and $s_n$ is the sum of the first $n$ terms. Finally, let $R_n = s-s_n$ be the remainder after summing the first $n$ terms. Then it is shown in the above link that $R_n$ lies between \(\displaystyle \int_n^\infty \!\!\!f(x)\,dx\) and \(\displaystyle \int_{n+1}^\infty \!\!\!f(x)\,dx.\) It suggests that a good estimate for $s$ would be to approximate $R_n$ by the midpoint of those two bounds. In other words, you should use \(\displaystyle \frac12\biggl(\int_n^\infty \!\!\!f(x)\,dx + \int_{n+1}^\infty f(x)\,dx\biggr)\) as an approximation for $R_n$, getting $$s \approx s_n + \frac12\biggl(\int_n^\infty \!\!\!f(x)\,dx + \int_{n+1}^\infty f(x)\,dx\biggr).$$ In your problem, you should take $f(x) = \dfrac1{x^{1.01}}$ and $n=1$. Then $s_1 = 1$ and $$s \approx 1 + \frac12\biggl(\int_1^\infty \frac1{x^{1.01}}\,dx + \int_2^\infty \frac1{x^{1.01}}\,dx\biggr).$$ If you do that and evaluate the integrals correctly, you should find that $s\approx 100.65$, as the question suggests.