Answer
Convergent with sum $\dfrac{49}{3}$
Work Step by Step
We are given the series:
$\sum_{n=1}^{\infty} \dfrac{7^{n+1}}{10^n}$.
Rewrite the series:
$a_n=\sum_{n=1}^{\infty} 7\left(\dfrac{7}{10}\right)^n$.
A partial sum for $m$ terms can be written:
$S_m=\sum_{n=1}^{m} 7\left(\dfrac{7}{10}\right)^n=7\left(\dfrac{7}{10}\right)^1+7\left(\dfrac{7}{10}\right)^2+...+7\left(\dfrac{7}{10}\right)^m$
$=7\left(\dfrac{7}{10}\right)^1\cdot\dfrac{1-\left(\dfrac{7}{10}\right)^m}{1-\dfrac{7}{10}}$
$=7\left(\dfrac{7}{10}\right)\left(\dfrac{10}{3}\right)\cdot \left[1-\left(\dfrac{7}{10}\right)^m\right]$
$=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^m\right]$
Determine at least 10 partial sums:
$S_1=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^1\right]=4.9$
$S_2=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^2\right]\approx 8.33$
$S_3=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^3\right]\approx 10.731$
$S_4=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^4\right]\approx 12.4117$
$S_5=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^5\right]\approx 13.58819$
$S_{10}=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^{10}\right]\approx 15.87198$
$S_{25}=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^{25}\right]\approx 16.33114$
$S_{50}=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^{50}\right]\approx 16.33333$
$S_{100}=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^{100}\right]\approx 16.33333$
$S_{500}=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^{500}\right]\approx 16.33333$
$S_{1000}=\dfrac{49}{3}\left[1-\left(\dfrac{7}{10}\right)^{1000}\right]\approx 16.33333$
Graph both the sequence of terms (in blue color) and the sequence of partial sums (in red color).
When $m\rightarrow \infty$, $\left(\dfrac{7}{10}\right)^m\rightarrow 0$, so $S_m\rightarrow \dfrac{49}{3}$, therefore $a_n\rightarrow \dfrac{49}{3}$. Therefore the series is convergent, its sum being $\dfrac{49}{3}$.
