Mathematical Reflections 2013, Issue 5 - Problem U278

Problem:
Evaluate $$\lim_{n \to \infty} \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!}.$$

Proposed by Dorin Andrica.

Solution:
Observe that for any $n,k \geq 1$, it holds $n+1 \leq kn+1 \leq n(k+1)$. Therefore,
$$1+\sum_{k=1}^\infty \dfrac{1}{n(k+1)!} \leq \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!} \leq 1+\sum_{k=1}^\infty \dfrac{1}{(n+1)k!},$$ i.e. $$1+\dfrac{e-2}{n} \leq \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!} \leq 1+\dfrac{e-1}{n+1}.$$
By the Squeeze Theorem, we have
$$1 \leq \lim_{n \to \infty} \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!} \leq 1,$$ hence $\displaystyle \lim_{n \to \infty} \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!}=1$.