Can someone explain the root test and when to use it?
Matthew Barrera 
Can someone dumb down and explain the root test for the convergence of a series. Also How to identify when to use it and how?
my intuition tells me to use the root test on this problem but I don't have the understanding to apply it
$$\sum_{k=1}^{\infty}\Big(\frac{k+1}{2k+3}\Big)^k$$
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$\begingroup$You need to take k-th root of sum's argument and calculate the limit:
$$L=\lim_{k \to \infty}\sqrt[k]{\left(\frac{k+1}{2k+3}\right)^k}= \lim_{k \to \infty} \frac{k+1}{2k+3}=\frac{1}{2}$$
Based on the value of $L$ we can determine if the sequence is convergent ($L<1$) or divergent ($L>1$). If the limit turns out to be equal to $1$, we cannot determine if the sequence converges and other test needs to be used.
In our case $L=\frac{1}{2}<1$, which means the sequence is convergent.
$\endgroup$ $\begingroup$We have that
$$\sqrt[k]{a_k}=\frac{k+1}{2k+3}=\frac{1+\frac1k}{2+\frac3k} \to \frac{1+0}{2+0}=\frac12<1$$
therefore the series converges.
As an alternative we have
$$\Big(\frac{k+1}{2k+3}\Big)^k=\frac1{2^k}\Big(\frac{2k+2}{2k+3}\Big)^k\le \frac1{2^k}$$
and conclude by direct comparison test.
$\endgroup$ $\begingroup$Let the index of summation be $k$.
The root test is easy to apply, as in your example, when there are expressions raised to the$k$-th power. Taking the$k$-th root just means dividing the exponents by $k$.
Note that if$k$ is raised to a fixed power, such as $k^a$, that this will not affect the radius of convergence, since$k^{a/k} \to 1$as $k \to \infty$. This might affect whether or not the sum converges at the boundaries of the region of convergence.
If the expression has factorials, the ratio test might be easier since the ratio of consecutive terms of an expression involving factorials is often easy to compute.
If you want to use the root test on expressions involving factorials, just use Stirling to get
$\begin{array}\\ ((ak+b)!)^{1/k} &\approx ((\sqrt{2\pi(ak+b)}(\frac{ak+b}{e})^{ak+b})^{1/k}\\ &\to (\frac{ak+b}{e})^{a+b/k}\\ &\to (\frac{ak+b}{e})^{a}\\ &\to (\frac{ak}{e})^{a}(1+b/(ak))^{a}\\ &\to (\frac{ak}{e})^{a}\\ \end{array} $
This, by itself, will diverge. However, there are usually other terms involving powers of $k$that will cancel out so the result will depend on the coefficients of the various factorial terms.
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