convergence of $\sum_{n=1}^{\infty} \frac{a^n}{n!}$

Question

How can i check convergence of $$\sum_{n=1}^{\infty} \frac{a^n}{n!}$$ ?

I tried some of the tests for checking if it's convergent , but it's does not work.

Answer 1

HINT: This series converges to $e^a - 1$.

Hence...

Proof of the convergence

We have different ways to prove this. The first one is to prove that $e^x$ minus the above series is an $\epsilon$:

We have to show that for any $\epsilon\gt0$ and $M\gt0$ there is an $N$ so that if $n\ge N$ and $|x|\le M$, then $$ \left|e^x-\sum_{k=0}^n\frac{x^k}{k!}\right|\le\epsilon $$ We can achieve this by showing $$ \sum_{k=n+1}^\infty\frac{|x|^k}{k!}\le\epsilon $$ For $k\ge2M$, $\frac{M^k}{k!}=\frac{M^{2M}}{(2M)!}\overbrace{\frac{M}{2M+1}\frac{M}{2M+2}\cdots\frac{M}{k}}^{k-2M\text{ terms}}\le\frac{M^{2M}}{(2M)!}\frac1{2^{k-2M}}=\frac{(2M)^{2M}}{(2M)!}2^{-k}$.

By choosing $N\ge\log_2\left(\frac1\epsilon\frac{(2M)^{2M}}{(2M)!}\right)$, we get for $n\ge N$ and $|x|\le M$, $$ \begin{align} \sum_{k=n+1}^\infty\frac{|x|^k}{k!} &\le\sum_{k=n+1}^\infty\frac{M^k}{k!}\\ &\le\sum_{k=n+1}^\infty\frac{(2M)^{2M}}{(2M)!}2^{-k}\\ &=\frac{(2M)^{2M}}{(2M)!}2^{-n}\\[9pt] &\le\epsilon \end{align} $$

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We can also use the error term $e^x-P_n(x)=\dfrac{e^\xi x^{n+1}}{(n+1)!}$ for some $\xi$ between $0$ and $x$. Given an $\epsilon\gt0$ and $M\gt0$, we want to find an $N$ so that if $n\ge N$ and $|x|\le M$, then $$ \frac{e^\xi x^{n+1}}{(n+1)!}\le\epsilon $$ For $k\ge2M$, $\frac{M^k}{k!}=\frac{M^{2M}}{(2M)!}\overbrace{\frac{M}{2M+1}\frac{M}{2M+2}\cdots\frac{M}{k}}^{k-2M\text{ terms}}\le\frac{M^{2M}}{(2M)!}\frac1{2^{k-2M}}=\frac{(2M)^{2M}}{(2M)!}2^{-k}$.

By choosing $N\ge\log_2\left(\frac{e^M}\epsilon\frac{(2M)^{2M}}{(2M)!}\right)$, we get for $n\ge N$ and $|x|\le M$ and $\xi$ between $0$ and $x$, $$ \begin{align} \frac{e^\xi x^{n+1}}{(n+1)!} &\le\frac{e^MM^{n+1}}{(n+1)!}\\ &\le e^M\frac{(2M)^{2M}}{(2M)!}2^{-n-1}\\[9pt] &\le\epsilon \end{align} $$

Second way - Lagrange Reminders

We have an amazing thing called Lagrange remainders. They basically tell us the difference between our function and it's Taylor polynomial. In general, we have

$$R_n(x)=|f(x)-P_n(x)|$$

where $P_n(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k$. Since it follows that

$$R_n(a)=0\\R_n'(a)=0\\R_n''(a)=0\\\vdots\\R_n^{(n)}(a)=0\\R_n^{(n+1)}(a)=|f^{(n+1)}(a)|$$

Thus,

$$R_n^{(n+1)}(x)\le|f^{(n+1)}(c)|$$

for some $c$ in our radius of convergence. It thus follows by integrating a few times that

$$R_n(x)\le\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|$$

One can then see that as $n\to\infty$, we have

$$|f(x)-P(x)|\le\lim_{n\to\infty}\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|$$

and if $\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\to0$ for any $x,c$ within the a given domain, then the power series will equal the original function over that domain.

See if you can show that for any $x,c\in\mathbb R$,

$$\lim_{n\to\infty}\left|\frac{e^c}{(n+1)!}x^{n+1}\right|=0$$

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On a side note, Lagrange remainder also shows us how well we approximate something when using a power series. For example, if I wanted to calculate $e$ out 5 places accurately,

$$R_n(x)=\left|e^x-\sum_{k=0}^n\frac{x^n}{n!}\right|\le0.000001$$

It's easy enough to solve, since

$$R_n(x)\le\left|e\frac{x^{n+1}}{(n+1)!}\right|\le\left|3\frac{x^{n+1}}{(n+1)!}\right|$$

Our particular case is $x=1$, and thus it suffices to solve

$$\frac3{(n+1)!}<0.000001$$

Which is easily done with a few checks to give $n\le8$. Thus,

$$e=\pm0.000001+\sum_{k=0}^8\frac1{k!}$$

Corollary

You can use the same methods for your series:

$$\sum_{n = 1}^{+\infty} \frac{a^n}{n!} \leq e^a - 1$$

Answer 2

Hint: You can use the ratio test. Compute $$\lim_{n\rightarrow\infty}\frac{a^{n+1}/(n+1)!}{a^n/n!}$$