Editing Transcendental number (section)

====Lemma 2====
''For sufficiently large {{mvar|k}}, <math>\left| \tfrac{Q}{k!} \right| <1</math>.''

'''Proof.''' Note that

<math display=block>\begin{align} 
  f_k e^{-x} &= x^{k} \left[ (x-1)(x-2) \cdots (x-n) \right]^{k+1} e^{-x}\\
  &= \left (x(x-1)\cdots(x-n) \right)^k \cdot \left( (x-1) \cdots (x-n) e^{-x} \right) \\
  &= u(x)^k \cdot v(x)
\end{align}</math>

where {{math|''u''(''x''), ''v''(''x'')}} are [[Continuous function|continuous functions]] of {{mvar|x}} for all {{mvar|x}}, so are bounded on the interval {{math|[0, ''n'']}}. That is, there are constants {{math|''G'', ''H'' > 0}} such that

<math display=block>\ \left| f_k e^{-x} \right| \leq |u(x)|^k \cdot |v(x)| < G^k H \quad \text{ for } 0 \leq x \leq n ~.</math>

So each of those integrals composing {{mvar|Q}} is bounded, the worst case being

<math display=block>\left| \int_{0}^{n} f_{k} e^{-x}\ \mathrm{d}\ x \right| \leq \int_{0}^{n} \left| f_{k} e^{-x} \right| \ \mathrm{d}\ x \leq \int_{0}^{n}G^k H\ \mathrm{d}\ x = n G^k H ~.</math>

It is now possible to bound the sum {{mvar|Q}} as well:

<math display=block> |Q| < G^{k} \cdot n H \left( |c_1|e+|c_2|e^2 + \cdots+|c_n|e^{n} \right) = G^k \cdot M\ ,</math>

where {{mvar|M}} is a constant not depending on {{mvar|k}}. It follows that

<math display=block>\ \left| \frac{Q}{k!} \right| < M \cdot \frac{G^k}{k!} \to 0 \quad \text{ as } k \to \infty\ ,</math>

finishing the proof of this lemma.