Convergences of series - Geometric series & quotient rule

Question

I want to check if the following series converge or not.

\begin{align*}&1. \ \ \ \sum_{n=0}^{\infty}\frac{4\cdot 4^n+10\cdot 3^n}{5^n} \\ &2. \ \ \ \sum_{n=0}^{\infty}\frac{7^{n+3}}{(n+9)!}\end{align*}

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I have done the following :

1. We can write this sum as a sum of two sums and use then the geometric series : \begin{align*}\sum_{n=0}^{\infty}\frac{4\cdot 4^n+10\cdot 3^n}{5^n}&=\sum_{n=0}^{\infty}\frac{4\cdot 4^n}{5^n}+\sum_{n=0}^{\infty}\frac{10\cdot 3^n}{5^n}\\ & =4\cdot\sum_{n=0}^{\infty}\frac{4^n}{5^n}+10\cdot\sum_{n=0}^{\infty}\frac{ 3^n}{5^n} \\ & =4\cdot\sum_{n=0}^{\infty}\left (\frac{4}{5}\right )^n+10\cdot\sum_{n=0}^{\infty}\left (\frac{ 3}{5}\right )^n\end{align*} Since $\left |\frac{4}{5}\right |<1$ and $\left |\frac{3}{5}\right |<1$ we have that both of them converge, since these are geometric series.

We can also calculate the limit : \begin{align*}\sum_{n=0}^{\infty}\frac{4\cdot 4^n+10\cdot 3^n}{5^n}& =4\cdot\sum_{n=0}^{\infty}\left (\frac{4}{5}\right )^n+10\cdot\sum_{n=0}^{\infty}\left (\frac{ 3}{5}\right )^n\\ & =4\cdot\frac{1}{1-\frac{4}{5}}+10\cdot\frac{1}{1-\frac{3}{5}}\\ & =4\cdot 5+10\cdot\frac{5}{2}\\ & =20+25 \\ & = 45\end{align*}

2. When we have factorials and exponentials, then the quotient rule is very likely to be useful.

We have that \begin{equation*}a_n=\frac{7^{n+3}}{(n+9)!} \Rightarrow a_{n+1}=\frac{7^{n+1+3}}{(n+1+9)!}=\frac{7^{n+4}}{(n+10)!}\end{equation*} Then we have that \begin{align*}\left |\frac{a_{n+1}}{a_n}\right |&=\left |\frac{\frac{7^{n+4}}{(n+10)!}}{\frac{7^{n+3}}{(n+9)!}}\right |=\frac{\frac{7^{n+4}}{(n+10)!}}{\frac{7^{n+3}}{(n+9)!}}=\frac{7^{n+4}}{(n+10)!}\cdot \frac{(n+9)!}{7^{n+3}}=\frac{7\cdot 7^{n+3}}{(n+9)!\cdot (n+10)}\cdot \frac{(n+9)!}{7^{n+3}}\\ & =\frac{7}{(n+10)}<1\end{align*} So we get that the series converges.

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Is everything correct ?

Answer 1

Yes, everything is correct. Strictly speaking, to ascertain the convergence of the first series, we should not start with $\sum_{n=0}^\infty$ and start splitting the series into disjoint parts because, strictly speaking, we don’t yet know if that makes sense. But, it doesn’t really matter with a simple case like this.

With a substitution $m=n+9$, the last series is: $$\frac{1}{7^6}\sum_{m=9}^\infty\frac{7^m}{m!}=\frac{1}{7^6}\left(e^7-1-7-\frac{49}{2}-\cdots-\frac{7^8}{8!}\right)$$Which you can simplify if you wish.