# Show that the limit of a series is 0

Show $\lim\limits_{t \to \infty} e^{-t}\sum_{n=0}^{\infty}a_n\frac{t^n}{n!}=0$ if the sequence $\{a_n\}$ converges to limit 0.

So, If I could show that $\lim\limits_{t \to \infty} \sum_{n=0}^{\infty}a_n\frac{t^n}{n!}$ is finite then I am done as $\lim\limits_{t \to \infty} e^{-t}=0$. Now, $e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$ converges by ratio test. How to show $\lim\limits_{t \to \infty} \sum_{n=0}^{\infty}a_n\frac{t^n}{n!}$ is convergent?

• Do you mean $t^n$? – asdq Feb 12 '18 at 12:40
• I didn't get you. Could you be a bit more specific? Where do you suggest a correction? – Dom Feb 12 '18 at 12:46
• You wrote $t_n$ but I don't know what $t_n$ is supposed to be. – asdq Feb 12 '18 at 12:48
• Yes, I got it now. Thanks! I edited the question. – Dom Feb 12 '18 at 12:51

If $a_n\to 0$ then for every $\varepsilon > 0$ there is some $N=N(\varepsilon)$ ensuring $|a_n|\leq \varepsilon$ for any $n\geq N$.
Let $M=M(\varepsilon)$ the maximum of $|a_n|$ for $n\in[0,N]$. We have $$\left|\sum_{n\geq 0}a_n\frac{t^n}{n!}\right| \leq M\sum_{n=0}^{N}\frac{t^n}{n!}+\varepsilon e^t$$ hence $$\left|\lim_{t\to +\infty} e^{-t}\sum_{n\geq 0}a_n\frac{t^n}{n!}\right|\leq \varepsilon +M\lim_{t\to +\infty}\sum_{n=0}^{N}\frac{t^n e^{-t}}{n!}\leq \varepsilon$$ and since $\varepsilon$ is arbitrary, the wanted limit is zero.

The power series certainly converges for every $t$, because if $A=\max\{|a_n|:n\in\mathbb{N}\}$, then $$\left|\frac{a_nt^n}{n!}\right|\le A\frac{t^n}{n!}$$

Unfortunately for you, the series converges to a function $f(t)$ that can have infinite limit at $\infty$. For instance, if $a_n=1/(n+1)$, the series becomes $$\sum_{n=0}^\infty\frac{t^n}{(n+1)!}=\frac{1}{t}\sum_{n=0}^\infty\frac{t^{n+1}}{(n+1)!}=\frac{1}{t}(e^t-1)$$ (for $t\ne0$) and $\lim_{t\to\infty}(e^t-1)/t=\infty$.

On the other hand, $$\lim_{t\to\infty}\frac{(e^t-1)/t}{e^t}=0$$ so the statement you want to prove holds for this case nonetheless.

For completeness, here'a an argument (similar to Jack's, of course).

Let's use the fact that $\lim_{n\to\infty}a_n=0$. Then, for every $\varepsilon>0$, there exists $N$ with $|a_n|<\varepsilon$. Thus we have (working for $t>0$) $$|f(t)|\le\sum_{n=0}^N\frac{|a_n|}{n!}t^n+\varepsilon\sum_{n>N}\frac{t^n}{n!}= \sum_{n=0}^N\frac{|a_n|}{n!}t^n-\varepsilon\sum_{n=0}^N\frac{t^n}{n!} +\varepsilon\sum_{n\ge0}\frac{t^n}{n!}=p(t)+\varepsilon e^t$$ where $p(t)$ is a polynomial of degree at most $N$.

Since $\lim_{t\to\infty}\dfrac{p(t)}{e^t}=0$, you should be able to conclude.