Find the interval of convergence and sum of the geometric series with ratio -5x

Key concept: This is a geometric series in disguise. Once you spot the common ratio, the convergence test and the sum formula both drop out pretty quickly.

Step by step

We have $\sum_{n=1}^{\infty}(-5)^n x^n.$

Step 1: Rewrite it as a geometric series

a typical term is $(-5)^n x^n = (-5x)^n.$ So the series becomes $\sum_{n=1}^{\infty}(-5x)^n.$ This is geometric with common ratio $r=-5x.$

Step 2: Use the convergence condition

A geometric series converges when $|r|<1.$ So here, $|-5x|<1,$ which gives $|x|<\frac15.$ Therefore the interval of convergence is $\left(-\frac15,\frac15\right).$

Step 3: Find the sum

Since the series starts at $n=1$, $\sum_{n=1}^{\infty} r^n = \frac{r}{1-r}$ for $|r|<1$. Substitute $r=-5x$: $\sum_{n=1}^{\infty}(-5x)^n = \frac{-5x}{1-(-5x)} = \frac{-5x}{1+5x}.$

Final result

• Converges for: $-\frac15<x<\frac15$
• Sum: $\frac{-5x}{1+5x}$

Pitfall alert

The most common mistake is using the wrong geometric-series formula. Because the sum starts at $n=1$, you need $\frac{r}{1-r}$, not $\frac{1}{1-r}$. Another slip is checking only the algebraic inequality $-1<-5x<1$ and forgetting to rewrite it as an absolute-value condition first; both are fine, but the sign flip must be handled carefully.

Try different conditions

If the series had started at $n=0$, the sum would be $\sum_{n=0}^{\infty}(-5x)^n=\frac{1}{1+5x},$ with the same convergence interval $|x|<\frac15$. If the ratio were changed to $ax$ instead of $-5x$, the convergence rule would become $|ax|<1$, and the sum would be $\frac{ax}{1-ax}$ for a series starting at $n=1$.

Further reading

geometric series, radius of convergence, common ratio