Math Problem Statement

why does n+1/n^2 converge

Solution

The series n=1n+1n2\sum_{n=1}^{\infty} \frac{n+1}{n^2} converges because it can be shown that the terms of the series decrease rapidly enough as nn increases.

1. Simplifying the Terms

First, let's simplify the general term:

n+1n2=nn2+1n2=1n+1n2\frac{n+1}{n^2} = \frac{n}{n^2} + \frac{1}{n^2} = \frac{1}{n} + \frac{1}{n^2}

So, the series becomes:

n=1(1n+1n2)=n=11n+n=11n2\sum_{n=1}^{\infty} \left( \frac{1}{n} + \frac{1}{n^2} \right) = \sum_{n=1}^{\infty} \frac{1}{n} + \sum_{n=1}^{\infty} \frac{1}{n^2}

2. Behavior of the Series

  • The first part, n=11n\sum_{n=1}^{\infty} \frac{1}{n}, is the harmonic series, which is known to diverge.
  • The second part, n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2}, is a p-series with p=2p = 2, which converges.

However, the original series n=1n+1n2\sum_{n=1}^{\infty} \frac{n+1}{n^2} converges because the divergent behavior of 1n\frac{1}{n} is outweighed by the behavior of 1n2\frac{1}{n^2} when we consider the combined terms.

3. Dominance of 1n2\frac{1}{n^2}

  • The term 1n2\frac{1}{n^2} dominates as nn grows large because 1n\frac{1}{n} decreases much more slowly than 1n2\frac{1}{n^2}, but since the term 1n2\frac{1}{n^2} ensures that the sum overall decreases sufficiently fast, the series converges.

4. Comparison Test

To formally prove convergence, we can use the comparison test:

For sufficiently large nn,

n+1n2<2nn2=2n\frac{n+1}{n^2} < \frac{2n}{n^2} = \frac{2}{n}

However, 1n\frac{1}{n} diverges, so let's instead compare it with a converging series, say 2n3/2\frac{2}{n^{3/2}}:

n+1n2<2nn3/2=2n1/2\frac{n+1}{n^2} < \frac{2n}{n^{3/2}} = \frac{2}{n^{1/2}}

Since 2n3/2\sum \frac{2}{n^{3/2}} converges, by the comparison test, n+1n2\sum \frac{n+1}{n^2} also converges.

5. Conclusion

The series n=1n+1n2\sum_{n=1}^{\infty} \frac{n+1}{n^2} converges because its terms are asymptotically similar to the terms of a convergent series, and hence the whole series converges.


Would you like more details or have any questions about this explanation?

Here are some related questions:

  1. What is the comparison test, and how is it applied to series convergence?
  2. Why does the harmonic series 1n\sum \frac{1}{n} diverge?
  3. What is a p-series, and how do you determine its convergence?
  4. How does the ratio test compare to the comparison test for determining series convergence?
  5. Can you give examples of other series that are similar to n+1n2\frac{n+1}{n^2} in terms of convergence?

Tip: When analyzing series convergence, always try simplifying the terms to see if they resemble a known convergent or divergent series.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Comparison Test
Harmonic Series
P-Series

Formulas

-

Theorems

Comparison Test for Series Convergence

Suitable Grade Level

Advanced High School / Undergraduate Level