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There was much discussion on Math SO why

$$\lim_{n\to\infty} \frac{\alpha^n}{n!} = 0$$

when $\alpha > 1$. What would be the strategy for computing:

$$\sum_{n=0}^{\infty} \frac{\alpha^n}{n!}$$

given that it's convergence is easlity proven using ratio test.

Denis Kulagin
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3 Answers3

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The summatory you are proposing is the represantion of $e^{\alpha}$. This is obtained using Taylor Series : https://en.wikipedia.org/wiki/Taylor_series

$$e^\alpha = \sum^{\infty}_{n=0} \frac{\alpha^n}{n!} $$

As your summatory start with $n=1$ then it would be: $$e^{\alpha }-1 = \sum^{\infty}_{n=1} \frac{\alpha^{n}}{n!}$$, since:

$$ e^\alpha = \sum^{\infty}_{n=0} \frac{\alpha^n}{n!}=\frac{\alpha^{0}}{0!} + \sum^{\infty}_{n=1} \frac{\alpha^{n}}{n!} = 1 +\sum^{\infty}_{n=1} \frac{\alpha^{n}}{n!} $$

Ivan
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  • As your summatory start with $n=1$ then it would be: $e^{\alpha }-1 = \sum^{\infty}{n=1} \frac{\alpha^{n}}{n!}$, since $ e^\alpha = \frac{\alpha^{0}}{0!} + \sum^{\infty}{n=1} \frac{\alpha^{n}}{n!} = 1 +\sum^{\infty}_{n=1} \frac{\alpha^{n}}{n!} $ – Ivan Apr 20 '17 at 15:51
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The power series $\sum_{n=0}^\infty \frac{x^n}{n!}$ converges for all $x$. If we call it by $f(x)$, notice that $$ f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n!} = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{m=0}^\infty \frac{x^{m}}{m!} = f(x) $$ Also, $f(0) = 1$. These two properties characterize the exponential function, by uniqueness of solutions to ordinary differential equations. Therefore $f(x) = e^x$ for all $x$.

Matthew Leingang
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  • There are several characterizations of the exponential function. The power series representation is one. Others include the limit definition, the inverse of $\int_1^x \frac1t,dt$, and the Lebesgue measureable function that satisfies the functional equation $f(x+y)=f(x)f(y)$, with $f(1)=e$. The ODE that you discuss is another one yet. The point is that we have the series to begin and can take that as The Characterization. And we are done. – Mark Viola Apr 20 '17 at 16:47
  • @Dr.MV I agree with everything you said except the sentence "The point is..." If the student is trying to show that the series converges to $e^\alpha$ and answers “by definition”, and is working within a framework where that is not the definition of the exponential function, then that answer doesn't help. – Matthew Leingang Apr 20 '17 at 16:56
  • @Dr.MV: When I was taking my quals a satirical version was distributed at the same time with the question: "Define for a topological space $X$ and natural number $n$ the $n$th homotopy group $\pi_n(X)$. Compute $\pi_{14}(S^{10})$." A clever person responded that the best way to answer the question was to define $\pi_n(X)$ to be zero. Then $\pi_{14}(S^{10})$ is easy to calculate. – Matthew Leingang Apr 20 '17 at 16:59
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As supplemental to the other answers, and as I feel like that real powers should be defined before we get into Taylor series or differential equations, here is a direct approach using just the binomial theorem.

The usual definition of $e$ is $e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n$. Through this we can see that $$e^x=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{xn}=\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n,\quad x\in\mathbb{R}$$

So this is how we define $e^x$.

Using the binomial theorem, we may write $$\left(1+\frac{x}{n}\right)^n=\sum_{k=0}^n\frac{\binom{n}{k}}{n^k}x^k=\sum_{k=0}^n\frac{n(n-1)..(n-k+1)}{n^k}\frac{x^k}{k!}\leq\sum_{k=0}^n\frac{x^k}{k!}$$

as $(n-j)/n\leq 1$. Hence $$e^x\leq \sum_{k=0}^\infty\frac{x^k}{k!}$$

On the other hand, assuming that $x\geq 0$ we have for $n\geq m\geq 2$

$$\sum_{k=0}^m\frac{n(n-1)..(n-k+1)}{n^k}\frac{x^k}{k!}\leq\sum_{k=0}^n\frac{n(n-1)..(n-k+1)}{n^k}\frac{x^k}{k!}$$ Actually, we do not really have to assume positivity of $x$, just that $m$ is large enough and that $n,m$ are even, and then we can estimate the difference of the RHS and the LHS to be positive, and everything will still work for negative $x$ too.

Taking limits for $n\rightarrow\infty$ this gives $$\sum_{k=0}^m\frac{x^k}{k!}\leq e^x$$ for any $m>2$, hence $$\sum_{k=0}^\infty\frac{x^k}{k!}\leq e^x$$

Thus, we have proved that indeed the power series above is equal to $e^x$ $$\sum_{k=0}^\infty\frac{x^k}{k!}=e^x$$

dim-ask
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