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MathCalculus

Integral Test for Convergence: Does ∑ n/(3n² + 1) Converge?

The positive decreasing extension has a divergent logarithmic improper integral, so ∑ n/(3n² + 1) diverges by the integral test.

Question

Use the integral test to determine whether the series

∑n=1∞n3n2+1\sum_{n=1}^{\infty}\frac{n}{3n^2+1}

converges or diverges.

Answer

Let

f(x)=x3x2+1.f(x)=\frac{x}{3x^2+1}.

For x≥1x\ge1, f(x)>0f(x)>0 and ff is continuous. Its derivative is

f′(x)=1−3x2(3x2+1)2<0,f'(x)=\frac{1-3x^2}{(3x^2+1)^2}<0,

so ff is decreasing on [1,∞)[1,\infty). Also, f(n)=n/(3n2+1)f(n)=n/(3n^2+1), so the integral test applies.

Evaluate the corresponding improper integral:

∫1∞x3x2+1 dx=lim⁡b→∞16[ln⁡(3x2+1)]1b=lim⁡b→∞16(ln⁡(3b2+1)−ln⁡4)=∞.\begin{aligned} \int_1^{\infty}\frac{x}{3x^2+1}\,dx &=\lim_{b\to\infty}\frac16\left[\ln(3x^2+1)\right]_1^b\\ &=\lim_{b\to\infty}\frac16\left(\ln(3b^2+1)-\ln4\right)\\ &=\infty. \end{aligned}

Because the improper integral diverges, the integral test gives

∑n=1∞n3n2+1 diverges.\boxed{\displaystyle\sum_{n=1}^{\infty}\frac{n}{3n^2+1}\text{ diverges}.}

Evidence boundary

This conclusion uses the positive, continuous, decreasing extension f(x) = x/(3x² + 1) on [1, ∞). The integral test determines convergence or divergence; it does not assign a finite sum to this divergent series.

Sources

These references support the concepts and methods used in the explanation above.

Integral Test for Convergence: Series Answer | Verla