Determine the convergence interval of $\sum_{n=1}^{\infty} \left(\frac{x}{5}\right)^n$

Key takeaway: This one looks like a power series, but it is really a geometric series in disguise. Once you spot that, everything becomes quick: ratio, interval, and endpoints all line up neatly.

Consider

$\sum_{n=1}^{\infty} \frac{x^n}{5^n}=\sum_{n=1}^{\infty}\left(\frac{x}{5}\right)^n.$

This is a geometric series with common ratio

$r=\frac{x}{5}.$

Step 1: Convergence condition

A geometric series converges exactly when

$|r|<1.$

So we need

$\left|\frac{x}{5}\right|<1 \quad \Longrightarrow \quad |x|<5.$

That gives the open interval

$(-5,5).$

Step 2: Check the endpoints

• At $x=5$:

  $\sum_{n=1}^{\infty} 1$

  diverges.

• At $x=-5$:

  $\sum_{n=1}^{\infty} (-1)^n$

  also diverges, because the terms do not approach $0$.

Final answer

$\boxed{(-5,5)}$

and the radius of convergence is

$\boxed{R=5}.$

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Pitfalls the pros know 👇 Don’t be fooled by the alternating sign at $x=-5$. An alternating series still needs terms that go to zero. Here the terms are just $\pm 1$, so convergence fails right away. Also, endpoint checks matter even when the interior test is obvious.

What if the problem changes? If the series were $\sum \left(\frac{x-a}{5}\right)^n$, the center would shift to $x=a$ and the convergence interval would become $(a-5,a+5)$, with the same endpoint test logic.

Tags: geometric series, radius of convergence, endpoint test