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

Nov 7, 2015

Take the limit as $k \rightarrow \infty$ of a general term in the series divided by its preceding term, and apply the rules of the ratio test to see that the series converges.

#### Explanation:

In the ratio test, we check to see if the series $\sum {a}_{k}$ converges or diverges by examining the ratio ${a}_{k + 1} / {a}_{k}$ as $k \rightarrow \infty$

If ${\lim}_{k \rightarrow \infty} {a}_{k + 1} / {a}_{k} > 1$ then the series diverges.

If $0 \le {\lim}_{k \rightarrow \infty} {a}_{k + 1} / {a}_{k} < 1$ then the series converges.

If ${\lim}_{k \rightarrow \infty} {a}_{k + 1} / {a}_{k} = 1$ then the test does not work and a new method is needed.

In this case, a_k = ((2k)!)/k^(2k) so
lim_(krarroo)a_(k+1)/a_k = lim_(krarroo)(((2(k+1))!)/(k+1)^(2(k+1)))/(((2k)!)/k^(2k))
=>lim_(krarroo)a_(k+1)/a_k= lim_(krarroo)((2k+2)!)/((2k)!)*k^(2k)/(k+1)^(2(k+1))

((2k+2)!)/((2k)!) = (k+1)(k+2) and
${k}^{2 k} / {\left(k + 1\right)}^{2 \left(k + 1\right)} = {\left(\frac{k}{k + 1}\right)}^{2 k} \cdot \frac{1}{k + 1} ^ 2$

So, multiplying, we have
${\lim}_{k \rightarrow \infty} {a}_{k + 1} / {a}_{k} = {\lim}_{k \rightarrow \infty} \frac{k + 2}{k + 1} \cdot {\left(\frac{k}{k + 1}\right)}^{2 k}$

${\lim}_{k \rightarrow \infty} \frac{k + 2}{k + 1} = 1$ and
${\lim}_{k \rightarrow \infty} {\left(\frac{k}{k + 1}\right)}^{2 k} = {e}^{- 2}$ (see below for how to solve this part)

Then, multiplying gives us

${\lim}_{k \rightarrow \infty} {a}_{k + 1} / {a}_{k} = 1 \cdot {e}^{-} 2 = {e}^{-} 2$

As $0 \le {e}^{-} 2 < 1$, by the ratio test, the series sum((2k)!)/k^(2k) converges.

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To find ${\lim}_{k \rightarrow \infty} {\left(\frac{k}{k + 1}\right)}^{2 k}$ we will first look at ${\lim}_{k \rightarrow \infty} {\left(\frac{k}{k + 1}\right)}^{k}$

First, let's deal with the $k$ exponent:
lim_(krarroo)(k/(k+1))^k = lim_(krarroo)e^(ln((k/(k+1))^k)
$\implies {\lim}_{k \rightarrow \infty} {\left(\frac{k}{k + 1}\right)}^{k} = {\lim}_{k \rightarrow \infty} {e}^{k \ln \left(\frac{k}{k + 1}\right)} = {e}^{{\lim}_{k \rightarrow \infty} k \ln \left(\frac{k}{k + 1}\right)}$

Now, looking at ${\lim}_{k \rightarrow \infty} k \ln \left(\frac{k}{k + 1}\right)$, we will deal with the natural log using L'Hopital's rule.

${\lim}_{k \rightarrow \infty} k \ln \left(\frac{k}{k + 1}\right) = {\lim}_{k \rightarrow \infty} \ln \frac{\frac{k}{k + 1}}{\frac{1}{k}}$

As ${\lim}_{k \rightarrow \infty} \ln \left(\frac{k}{k + 1}\right) = {\lim}_{k \rightarrow \infty} \frac{1}{k} = 0$ we can apply L'Hopital's rule to obtain

${\lim}_{k \rightarrow \infty} \ln \frac{\frac{k}{k + 1}}{\frac{1}{k}} = {\lim}_{k \rightarrow \infty} \frac{\left(\frac{k + 1}{k}\right) \cdot \frac{k + 1 - k}{k + 1} ^ 2}{- \frac{1}{k} ^ 2}$

Simplifying, we get

${\lim}_{k \rightarrow \infty} k \ln \left(\frac{k}{k + 1}\right) = {\lim}_{k \rightarrow \infty} - \frac{k}{k + 1} = - 1$

Then we substitute back to obtain

${\lim}_{k \rightarrow \infty} {\left(\frac{k}{k + 1}\right)}^{k} = {e}^{-} 1$

Thus

${\lim}_{k a r r \infty} {\left(\frac{k}{k + 1}\right)}^{2 k} = {\lim}_{k \rightarrow \infty} {\left({\left(\frac{k}{k + 1}\right)}^{k}\right)}^{2} = {e}^{-} 2$