Does \(\sum \frac{n^{2n}}{(n+1)^{3n}}\) Converge?

Key concept: The exponent on $n$ makes this look intimidating, but the root test is the right tool. Once you take the $n$th root, the expression becomes much simpler.

Step by step

Step 1: Look at the $n$th root

Let

$a_n=\frac{n^{2n}}{(n+1)^{3n}}.$

Apply the root test:

$\sqrt[n]{a_n}=\frac{n^2}{(n+1)^3}.$

Step 2: Find the limit

$\lim_{n\to\infty}\frac{n^2}{(n+1)^3}=\lim_{n\to\infty}\frac{n^2}{n^3\left(1+\frac{1}{n}\right)^3}=\lim_{n\to\infty}\frac{1}{n\left(1+\frac{1}{n}\right)^3}=0.$

Step 3: Conclude from the root test

Since

$L=0<1,$

the series

$\sum_{n=1}^{\infty}\frac{n^{2n}}{(n+1)^{3n}}$

converges absolutely.

A small correction to a common thought

It is not enough to say "the series is smaller than $1/n$, so it converges." The harmonic series $\sum 1/n$ actually diverges. What matters is that the root test gives a limit strictly less than 1, which settles it immediately.

Pitfall alert

A frequent mistake is confusing "smaller than a divergent series" with convergence. Being less than $1/n$ does not prove anything, because $\sum 1/n$ diverges. Here the clean proof is the root test, not a comparison to the harmonic series.

Try different conditions

If the series were $\sum \frac{n^{2n}}{(n+1)^{2n}}$, then the $n$th root would be $\frac{n^2}{(n+1)^2}\to 1$, so the root test would be inconclusive. If the exponent in the denominator increased to $4n$, convergence would be even faster.

Further reading

root test, absolute convergence, exponential series