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Exercise 1515Calculus › Single-variable calculus › Series › Convergence testsL1Added 2 Nov 2026

The ratio test with a parameter

#ratio test#Stirling formula#critical case

Statement

For a>0a>0, study ∑n≥1n! annn\displaystyle\sum_{n\ge1}\frac{n!\,a^n}{n^n}.

Hint

Compute un+1un\frac{u_{n+1}}{u_n}. At the critical value of aa the ratio test fails: use Stirling's formula n!∼2πn (n/e)nn!\sim\sqrt{2\pi n}\,(n/e)^n.

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Solution

Ratio.
un+1un=a⋅(n+1) nn(n+1)n+1=a(1+1n)−n⟶ae.\frac{u_{n+1}}{u_n}=a\cdot\frac{(n+1)\,n^n}{(n+1)^{n+1}}=a\left(1+\frac1n\right)^{-n}\longrightarrow\frac ae.

Cases. For a<ea<e the series converges; for a>ea>e it diverges.

Critical case a=ea=e. By Stirling,
un=n! ennn∼2πn⟶+∞,u_n=\frac{n!\,e^n}{n^n}\sim\sqrt{2\pi n}\longrightarrow+\infty,
so the terms do not tend to 00: the series diverges.

Conclusion. The series converges iff 0<a<e0<a<e.