Convergence of Series Using Integral Test

In summary, the conversation discusses using the integral test to determine the convergence or divergence of a series, specifically the series \Sigma^{\infty}_{n=1}\frac{n^{k-1}}{n^{k}+c}, where k is a positive integer. One approach mentioned is to consider the integral \int^{/infty}_{1}\frac{x^{k-1}}{x^{k}+c} dx and use the properties of logarithms to determine divergence. Another suggestion is to use u-substitution to simplify the integral.
  • #1
clairez93
114
0

Homework Statement



Use the integral test to determine the convergence or divergence of the series.

[tex]\Sigma^{\infty}_{n=1}[/tex][tex]\frac{n^{k-1}}{n^{k}+c}[/tex] k is a positive integer

Homework Equations





The Attempt at a Solution



Consider:

[tex]\int^{/infty}_{1}[/tex][tex]\frac{x^{k-1}}{x^{k}+c} dx[/tex]

Not sure how to integrate this expression to determine if it converges or diverges.
 
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  • #2
This is more of a guess on my part, but since the degree of the polynomial on the denominator is only one greater than that of the numerator, the answer will be the form of natural log (i.e.[tex]\int \frac{x^0}{x^1} dx = \ln_|x| + C[/tex]). That means the integral is divergent -> series is divergent. You could also check for divergence with the limit comparison test, with a divergent series such as 1/n.
 
Last edited:
  • #3
clairez93 said:
Consider:

[tex]\int^{\infty}_{1}\;\frac{x^{k-1}}{x^{k}+c} dx[/tex]

Not sure how to integrate this expression to determine if it converges or diverges.

What u-substitution will make this integral easy?
 
  • #4
Try letting u be the denominator and make du look like the numerator in the integrand.
 

FAQ: Convergence of Series Using Integral Test

What is the Integral Test for series convergence?

The Integral Test is a method used to determine whether an infinite series converges or diverges. It states that if a series has a corresponding continuous, positive, and decreasing function, then the series and the integral of that function will either both converge or both diverge.

How do you use the Integral Test to determine convergence?

To use the Integral Test, you first need to find a continuous, positive, and decreasing function that represents the series. Then, you take the integral of that function and evaluate it from 1 to infinity. If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

Can the Integral Test be used for all series?

No, the Integral Test can only be used for series with positive terms. If a series has negative terms or alternating signs, the Integral Test cannot be applied. In those cases, other tests such as the Alternating Series Test or Ratio Test can be used.

What is the relationship between the Integral Test and the p-Series Test?

The Integral Test is a more general version of the p-Series Test. The p-Series Test is a special case of the Integral Test, where the corresponding function is f(x) = 1/x^p. So, if the integral of 1/x^p converges, then the series 1/n^p also converges, and vice versa.

Can the Integral Test be used to determine the sum of a series?

No, the Integral Test only tells you whether a series converges or diverges, but it does not give the actual sum of the series. To find the sum of a convergent series, you can use other methods such as the Geometric Series formula or the Summation formula.

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