Use the techniques of geometric series

[karush]

$\tiny{242.WS10.a}$
\begin{align*}
&\textsf{use the techniques of geometric series} \\
&-\textsf {telescoping series, p-series, n-th term } \\
&-\textsf{divergence test, integral test, comparison test,} \\
&-\textsf{limit comparison test,ratio test, root test, } \\
&-\textsf {absolute convergence, alternating series test}
\end{align*}\begin{align*}
\displaystyle
S_n&=\sum_{n=1}^{\infty} \frac{7^n}{n!}\\
&=
\end{align*}
$\textsf{not sure what test to use on this, was thinking ratio test?}$

 

[MarkFL]

If we are to simply evaluate the given series, consider the following Maclaurin series:

$$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$

Can you proceed?

 

[karush]

\begin{align*}
\displaystyle
S_n&=\sum_{n=0}^{\infty} \frac{7^n}{n!}\\
e^7 &=1+7+\frac{7^2}{2!}
+\frac{7^3}{3!}+\frac{7^4}{4!}+\cdots \\
e^7 &=1+7+\frac{49}{2}+\frac{343}{6}+\frac{2401}{24}+\cdots
\end{align*}
$\textsf{however this problem was given with }$ $n=1$

 

[MarkFL]

\begin{align*}
\displaystyle
S_n&=\sum_{n=0}^{\infty} \frac{7^n}{n!}\\
e^7 &=1+7+\frac{7^2}{2!}
+\frac{7^3}{3!}+\frac{7^4}{4!}+\cdots \\
e^7 &=1+7+\frac{49}{2}+\frac{343}{6}+\frac{2401}{24}+\cdots
\end{align*}
$\textsf{however this problem was given with }$ $n=1$

So, then I would write:

$$e^x-1=\sum_{k=1}^{\infty}\frac{x^k}{k!}$$ ;)

 

[Prove It]

$\tiny{242.WS10.a}$
\begin{align*}
&\textsf{use the techniques of geometric series} \\
&-\textsf {telescoping series, p-series, n-th term } \\
&-\textsf{divergence test, integral test, comparison test,} \\
&-\textsf{limit comparison test,ratio test, root test, } \\
&-\textsf {absolute convergence, alternating series test}
\end{align*}\begin{align*}
\displaystyle
S_n&=\sum_{n=1}^{\infty} \frac{7^n}{n!}\\
&=
\end{align*}
$\textsf{not sure what test to use on this, was thinking ratio test?}$

The ratio test should work fine :)