Exercise 397Calculus › Single-variable calculus › Series › Convergence testsL1Added 2 Nov 2026

The ratio test

#ratio test#d'Alembert#factorials

Statement

Decide the nature of
$\sum\frac{n!}{n^n},\qquad\sum\frac{2^n}{n!},\qquad\sum\frac{n^2}{2^n},\qquad\sum\frac{(2n)!}{(n!)^2}.$
Then show why the test is inconclusive for $\sum\frac1n$ and $\sum\frac1{n^2}$ .

Hint

If $\frac{u_{n+1}}{u_n}\to L$ , the series converges for $L<1$ and diverges for $L>1$ .

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Solution

1. $\dfrac{u_{n+1}}{u_n}=\dfrac{(n+1)\,n^n}{(n+1)^{n+1}}=\left(1+\frac1n\right)^{-n}\to\frac1e<1$ : converges.

2. $\dfrac{u_{n+1}}{u_n}=\dfrac2{n+1}\to0<1$ : converges.

3. $\dfrac{u_{n+1}}{u_n}=\dfrac12\left(1+\frac1n\right)^2\to\frac12<1$ : converges.

4. $\dfrac{u_{n+1}}{u_n}=\dfrac{(2n+2)(2n+1)}{(n+1)^2}\to4>1$ : diverges.

Inconclusive case. For $\frac1n$ and $\frac1{n^2}$ the ratio tends to $1$ in both cases, yet the first series diverges and the second converges. When $L=1$ another test is needed.

The ratio test

Statement

Solution

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