To evaluate the infinite series

$\sum_{n=1}^{\infty} \frac{7(-1)^n}{10^{2n}},$

let's break it down.

Step 1: Recognize the Form of the Series

This series can be rewritten as: $7 \sum_{n=1}^{\infty} \frac{(-1)^n}{10^{2n}}.$ Thus, we focus on the series $\sum_{n=1}^{\infty} \left(-\frac{1}{100}\right)^n.$

Step 2: Identify it as a Geometric Series

This is a geometric series of the form $\sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r},$ where $a = -\frac{1}{100}$ and $r = -\frac{1}{100}$.

Step 3: Evaluate the Series

Since our series starts from $n=1$, we need to adjust for that by subtracting the $n=0$ term: $\sum_{n=1}^{\infty} \left(-\frac{1}{100}\right)^n = \frac{\frac{-1}{100}}{1 - \left(-\frac{1}{100}\right)} = \frac{-\frac{1}{100}}{1 + \frac{1}{100}}.$

Simplify the denominator: $= \frac{-\frac{1}{100}}{\frac{101}{100}} = -\frac{1}{101}.$

Step 4: Multiply by 7

Now, we return to our original series: $7 \cdot -\frac{1}{101} = -\frac{7}{101}.$

Final Answer

$\sum_{n=1}^{\infty} \frac{7(-1)^n}{10^{2n}} = -\frac{7}{101}.$

Would you like a breakdown of any specific step?

1. How are geometric series evaluated in general?
2. What conditions make an infinite series convergent?
3. How would this change if the power was different?
4. Can this be generalized to other values of the base in the denominator?
5. How does altering the sign affect convergence?

Tip: For alternating series, check the ratio’s absolute value—this helps confirm convergence.