Question

Let $$ S(p)=\sum_{n=1}^{\infty} \frac{1}{n^{p}}, \quad p>1 $$ Show that $$ \frac{1}{(p-1)(N+1)^{p-1}}

Short answer

Question: Prove the following inequality for the Riemann zeta function when p > 1: $$ \frac{1}{(p-1)(N+1)^{p-1}}<S(p)-\sum_{n=1}^{N} \frac{1}{n^{p}}<\frac{1}{(p-1)N^{p-1}} . $$ Solution: We used Theorem 4.3.10 and integration to find upper and lower bounds for the subtracted part of the sum, resulting in the given inequality.

Step-by-step solution

Understanding Theorem 4.3.10 and applying it to the exercise

Theorem 4.3.10 gives an estimation of the integral \(\int_1^N \frac{1}{x^p} dx\) for p > 1. We can rewrite it as: $$ \frac{1}{(p-1)N^{p-1}} < \int_1^N \frac{1}{x^p} dx < \frac{1}{(p-1)(N-1)^{p-1}}. $$ We will use this to find bounds for the series in the given exercise.

Rewrite the subtracted part of the sum as an integral

We want to find an upper and a lower bound for \(S(p)-\sum_{n=1}^{N} \frac{1}{n^{p}}\). We can rewrite this as: $$ S(p)-\sum_{n=1}^{N} \frac{1}{n^{p}} = \sum_{n=N+1}^{\infty} \frac{1}{n^p} = \int_{N+1}^{\infty} \frac{1}{x^p} dx. $$ Now we can use Theorem 4.3.10 to find bounds for this integral.

Using Theorem 4.3.10 to find the upper and lower bounds

To find the bounds, we will compare the integral \(\int_{N+1}^{\infty} \frac{1}{x^p} dx\) with the integral \(\int_N^{\infty} \frac{1}{x^p} dx\). Using Theorem 4.3.10, we get: $$ \frac{1}{(p-1)N^{p-1}} < \int_N^{\infty} \frac{1}{x^p} dx < \frac{1}{(p-1)(N-1)^{p-1}}. $$ Now, we can use the fact that for \(p>1\), the function \(\frac{1}{x^p}\) is decreasing, and thus the integrals have the following relationship: $$ \int_{N+1}^{\infty} \frac{1}{x^p} dx < \int_N^{\infty} \frac{1}{x^p} dx. $$ From this, we get the following inequalities: $$ \frac{1}{(p-1)(N+1)^{p-1}} < \int_{N+1}^{\infty} \frac{1}{x^p} dx < \frac{1}{(p-1)N^{p-1}}. $$

Writing the final result

Since we have found the upper and lower bounds for the integral corresponding to \(S(p)-\sum_{n=1}^{N} \frac{1}{n^{p}}\), our final result is as follows: $$ \frac{1}{(p-1)(N+1)^{p-1}}<S(p)-\sum_{n=1}^{N} \frac{1}{n^{p}}<\frac{1}{(p-1)N^{p-1}} . $$

Key concepts

Improper Integrals

In calculus, an improper integral involves integration over an infinite interval or of an unbounded function. An improper integral looks at an infinite series from a continuous perspective. This allows us to understand sums that extend indefinitely, such as in the function presented as \[ \int_1^N \frac{1}{x^p} \, dx \]for \(p > 1\).

• This notation implies the integral bounds are extended to infinity or the function itself may not be bounded over the range of integration.
• In this exercise, improper integrals serve to estimate the tail end of the series \(S(p)\), providing upper and lower bounds for this portion of the sum.

By understanding and applying the nature of improper integrals, one can deduce important properties about infinite sums related to their convergence. Furthermore, these integrals help when direct computation of an infinite series is complex, simplifying it to an integral estimation.

Infinite Series

An infinite series sums up an infinite number of terms, usually following a specific rule or formula. Infinite series often needs careful treatment to understand whether they converge to a finite value. In this particular exercise:

• The series, \(S(p) = \sum_{n=1}^{\infty} \frac{1}{n^p}\), is examined where \(p > 1\).
• This series converges when \(p > 1\) due to the terms being the reciprocal of a power, which decreases rapidly enough to ensure convergence.

We analyze this series by relating it to an improper integral, which helps synthesize its behavior and determine bounds for its finite partial sums. Recognizing the convergence properties of such series is key to interpreting their behavior practically. In calculus, series convergence provides fundamental insights into analytic calculations and is an essential concept for further mathematical applications.

Integration Bounds

Integration bounds indicate the limits between which integration is performed. They play a critical role in defining the area of calculation in an integral or series:

• In this problem, understanding bounds is vital as it translates a sum of terms beyond a certain point into an improper integral.
• The bounds \(N+1\) to \(\infty\) for the integral \(\int_{N+1}^{\infty} \frac{1}{x^p} \, dx\) are directly related to the series \(S(p)\).

Evaluating these integration bounds helps determine precise approximations for the partial sums in the series. Moreover, these bounds act as a critical bridge that connects series convergence and improper integral calculus.