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Prove that the given series diverges by showing that the \(N^{\text {th }}\) partial sum satisfies \(S_{N} \geq k \cdot N\) for some positive constant \(k\). $$ \sum_{n=1}^{\infty} \frac{n}{\sqrt{n^{2}+1}} $$

Short Answer

Expert verified
The series diverges as \( S_N \geq \frac{1}{\sqrt{2}} \cdot N \), where \( \frac{1}{\sqrt{2}} \) is a positive constant.

Step by step solution

01

Find the General Term

The general term of the series is given by \( a_n = \frac{n}{\sqrt{n^2 + 1}} \).
02

Simplify the General Term

Observe that \( a_n = \frac{n}{\sqrt{n^2+1}} = \frac{1}{\sqrt{1+\frac{1}{n^2}}} \). As \( n \to \infty \), \( \frac{1}{n^2} \to 0 \), so \( a_n \approx 1 \).
03

Estimate the Partial Sum

Consider the \(N^{\text{th}}\) partial sum \( S_N = \sum_{n=1}^{N} \frac{n}{\sqrt{n^2+1}} \). Using our estimate from Step 2, for large \( n \), \( \frac{n}{\sqrt{n^2+1}} \approx 1 \).
04

Establish a Lower Bound

We know that \( \frac{n}{\sqrt{n^2+1}} > \frac{n}{\sqrt{n^2 + n^2}} = \frac{1}{\sqrt{2}} \). Thus, each term \( a_n > \frac{1}{\sqrt{2}} \).
05

Compute the Partial Sum Lower Bound

Then \( S_N = \sum_{n=1}^{N} a_n > \sum_{n=1}^{N} \frac{1}{\sqrt{2}} = \frac{N}{\sqrt{2}} \). Therefore, \( S_N \geq \frac{1}{\sqrt{2}} \cdot N \).
06

Conclusion on Divergence

Since \( S_N \geq \frac{1}{\sqrt{2}} \cdot N \) and \( \frac{1}{\sqrt{2}} \) is a positive constant, the series diverges. The partial sum grows without bounds as \( N \to \infty \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Sums
When we talk about the partial sum of a series, we mean the sum of the first few terms of that series. For example, when considering the series \( \sum_{n=1}^{\infty} \frac{n}{\sqrt{n^2+1}} \), the \( N^{\text{th}} \) partial sum is represented as \( S_N = \sum_{n=1}^{N} \frac{n}{\sqrt{n^2+1}} \). This is simply the sum of the first \( N \) terms of the series.
Partial sums are powerful because they help us explore the behavior of infinite series by simplifying the problem to a finite number of terms. Specifically, they allow us to check whether a series might converge to a specific value, or in the case of divergence, whether it increases without bound.
Understanding partial sums gives us a way to approach infinite series methodically, by breaking down the infinite into manageable parts. As we look further into the behavior of these sums, we gain insight into the series' overall behavior.
Infinite Series
An infinite series is simply a series that continues indefinitely, meaning there is no final term. Mathematically, it's represented as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) are the terms of the series. In our context, the series \( \sum_{n=1}^{\infty} \frac{n}{\sqrt{n^2 + 1}} \) demonstrates how these terms continue endlessly.
The beauty of infinite series is in their complexity and the profound mathematical concepts they reveal. Infinite series play essential roles in various branches of mathematics, including calculus and analysis, where they're used to define and approximate functions. Understanding an infinite series involves determining whether the total sum of the terms (the series itself) leads to a finite number, known as convergence, or grows unlimited, known as divergence.
By looking at the partial sums of an infinite series, we can start to discern whether the series converges or diverges. Such explorations of infinite series open doors to further mathematical investigations and applications.
Series Estimation
Estimating a series involves finding a close approximation of its terms or partial sums. In simpler terms, it's about getting a practical look at the series without exact calculations. In our exercise, we estimated the general term \( \frac{n}{\sqrt{n^2+1}} \) by simplifying it for large \( n \), showing \( a_n \approx 1 \) as \( n \to \infty \).
Estimation techniques help predict a series' behavior without digging into every calculation detail.
  • This process often simplifies complex expressions to reveal the series' underlying patterns.
  • With these patterns, we can form educated guesses about the series' long-term behavior.
Estimation is particularly useful in dealing with complex or non-standard series where direct computation wouldn't be feasible or straightforward. Understanding estimation gives us clearer insights and quick assessments on the convergence and divergence of series, making it a crucial tool in mathematical analysis.
Lower Bound for Series
In series analysis, establishing a lower bound involves finding a value that the series never decreases below. For the series \( \sum_{n=1}^{\infty} \frac{n}{\sqrt{n^2+1}} \), we established a lower bound by showing each term \( a_n > \frac{1}{\sqrt{2}} \).
A lower bound helps demonstrate divergence by providing evidence that the series' sum increases beyond any fixed point. Specifically:
  • In our exercise, the inequality \( S_N \geq \frac{1}{\sqrt{2}} \cdot N \) suggests that as \( N \) increases, the sum \( S_N \) also grows indefinitely.
  • This growth shows that the partial sums don't approach a fixed value, indicating divergence.
Using lower bounds offers a conclusive perspective on series analysis, especially in proving divergence. It confirms that the series behaves as predicted, giving confidence in the outcome. Conclusively, an established lower bound equips us with a solid argument to understand and explain the behavior of series comprehensively.

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