# How do you use the ratio test to test the convergence of the series ∑k/(3+k^2)  from k=1 to infinity?

Aug 14, 2018

Let:

${a}_{k} = \frac{k}{3 + {k}^{2}}$

and evaluate the ratio:

$\left\mid {a}_{k + 1} / {a}_{k} \right\mid = \left\mid \frac{\frac{k + 1}{3 + {\left(k + 1\right)}^{2}}}{\frac{k}{3 + {k}^{2}}} \right\mid$

$\left\mid {a}_{k + 1} / {a}_{k} \right\mid = \left(\frac{k + 1}{k}\right) \left(\frac{3 + {k}^{2}}{4 + 2 k + {k}^{2}}\right)$

We have that:

${\lim}_{k \to \infty} \left\mid {a}_{k + 1} / {a}_{k} \right\mid = 1$

so the ration test is in effect inconclusive to determine whether the series:

${\sum}_{k = 1}^{\infty} {a}_{k}$

is convergent.

However if we consider the harmonic series:

${\sum}_{k = 1}^{\infty} \frac{1}{k}$

which is divergent and we apply the limit comparison test, we can see that:

${\lim}_{k \to \infty} {a}_{k} / \left(\frac{1}{k}\right) = {\lim}_{k \to \infty} {k}^{2} / \left(3 + {k}^{2}\right) = 1$

so, as the limit is finite, the two series have the same character and we can conclude that;

${\sum}_{k = 1}^{\infty} {a}_{k}$

is divergent.