Question

What is the sum of the series infinity e n-1 -7(3/8)^n?

Answer 1

Let

$S_N = \displaystyle \sum_{n=1}^N -7 \left(\frac38\right)^n = -7\left(\dfrac38 + \dfrac{3^2}{8^2} + \dfrac{3^3}{8^3} + \cdots + \dfrac{3^N}{8^N}\right)$

Then

$\dfrac38 S_N = -7\left(\dfrac{3^2}{8^2} + \dfrac{3^3}{8^3} + \dfrac{3^4}{8^4} + \cdots + \dfrac{3^{N+1}}{8^{N+1}}\right)$

$S_N - \dfrac38 S_N = -7 \left(\dfrac38 - \dfrac{3^{N+1}}{8^{N+1}}\right)$

$\dfrac58 S_N = -\dfrac{21}8 \left(1 - \left(\dfrac38\right)^N\right)$

$S_N = -\dfrac{21}5 \left(1 - \left(\dfrac38\right)^N\right)$

As $N\to\infty$, the exponential term will converge to zero, so the infinite sum converges to

$\displaystyle \sum_{n=1}^\infty -7 \left(\frac38\right)^n = \lim_{n\to\infty} S_N = \boxed{-\dfrac{21}5}$