How do you test for convergence for #sum ln(n)/n^2# for n=1 to infinity?

Question

Emma Aldridge · Accepted Answer

Probably the best method is to use the integral test.  The function #f(x)=ln(x)/(x^2)# is positive and decreasing for #x\geq 2#.  The improper integral #\int_{2}^{\infty}f(x)\ dx=lim_{b->\infty}\int_{2}^{b}(ln(x))/(x^2)\ dx# can be shown to converge by direct integration (using integration-by-parts with #u=ln(x)#, #du=1/x dx#, #dv=x^[-2} dx#, and #v=-x^{-1}#): #lim_{b->\infty}\int_{2}^{b}(ln(x))/(x^2)\ dx# #=lim_{b->\infty}(-ln(b)/b+ln(2)/2)+\int_{2}^{b}x^{-2}\ dx# #=ln(2)/2-lim_{b->\infty}((b^{-1}-1/2))=(ln(2)+1)/2#. The integral test now implies that #\sum_{n=2}^{\infty}f(n)=\sum_{n=2}^{\infty}\frac{ln(n)}{n^2}# converges. Therefore, #\sum_{n=1}^{\infty}\frac{ln(n)}{n^2}# converges.

William Abdalla · Answer

undefined You can test for convergence of the series $ \sum \frac{\ln(n)}{n^2} $ using the integral test:

1. Integrate the function $ f(x) = \frac{\ln(x)}{x^2} $ from 1 to infinity.
2. If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

Performing the integral:
$$ \int_{1}^{\infty} \frac{\ln(x)}{x^2} \, dx $$

Integration by parts:
$$ u = \ln(x) $$
$$ du = \frac{1}{x} \, dx $$
$$ dv = \frac{1}{x^2} \, dx $$
$$ v = -\frac{1}{x} $$

Applying integration by parts formula:
$$ \int \frac{\ln(x)}{x^2} \, dx = -\frac{\ln(x)}{x} + \int \frac{1}{x^2} \, dx $$

$$ = -\frac{\ln(x)}{x} - \frac{1}{x} + C $$

Evaluate the integral from 1 to infinity:
$$ = \lim_{b \to \infty} \left( -\frac{\ln(b)}{b} - \frac{1}{b} - \left(-\frac{\ln(1)}{1} - \frac{1}{1}\right) \right) $$

$$ = \lim_{b \to \infty} \left( -\frac{\ln(b)}{b} - \frac{1}{b} - \left(0 - 1\right) \right) $$

$$ = \lim_{b \to \infty} \left( -\frac{\ln(b)}{b} - \frac{1}{b} + 1 \right) $$

As $ b $ approaches infinity, $ \frac{\ln(b)}{b} $ approaches 0 (since the logarithmic function grows slower than any power of $ b $), and $ \frac{1}{b} $ approaches 0. Therefore, the integral converges to 1.

Since the integral converges, by the integral test, the series $ \sum \frac{\ln(n)}{n^2} $ also converges.

Scarlett Allison · Answer

undefined To test for convergence of the series $ \sum \frac{\ln(n)}{n^2} $ from $ n = 1 $ to infinity, you can use the Integral Test. This test states that if $ f(x) $ is continuous, positive, and decreasing for all $ x \geq 1 $, then the series $ \sum_{n=1}^{\infty} f(n) $ converges if and only if the improper integral $ \int_{1}^{\infty} f(x) \, dx $ converges.

In this case, $ f(x) = \frac{\ln(x)}{x^2} $. To apply the Integral Test, we integrate $ f(x) $ from 1 to infinity:

$$ \int_{1}^{\infty} \frac{\ln(x)}{x^2} \, dx $$

This integral can be evaluated using integration by parts or other methods. After integration, if the result is a finite value, then the series $ \sum \frac{\ln(n)}{n^2} $ converges. If the integral diverges, then the series also diverges.