What is the interval of convergence of #sum_1^oo (3)/n^(x)#?

Question

What is the interval of convergence of #sum_1^oo  (3)/n^(x)#?

Christopher Agresta · Accepted Answer

#x > 1# The sum #sum_(n=1)^oo1/n^alpha# converges for #alpha > 1# so #sum_(n=1)^oo3/n^x# converges for #x > 1#

Cameron Alexander · Answer

undefined The interval of convergence for the series $ \sum_{n=1}^{\infty} \frac{3}{n^x} $ can be determined using the Ratio Test.

The Ratio Test states that if $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L $, then the series converges if $ L < 1 $ and diverges if $ L > 1 $. If $ L = 1 $, the test is inconclusive.

For the given series $ \frac{3}{n^x} $, $ a_n = \frac{3}{n^x} $ and $ a_{n+1} = \frac{3}{(n+1)^x} $.

Taking the ratio $ \frac{a_{n+1}}{a_n} $, we get:
$$ \frac{\frac{3}{(n+1)^x}}{\frac{3}{n^x}} = \frac{n^x}{(n+1)^x} $$

Simplifying, we have:
$$ \frac{n^x}{(n+1)^x} = \left( \frac{n}{n+1} \right)^x $$

As $ n $ approaches infinity, $ \frac{n}{n+1} $ approaches 1, so $ \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^x = 1^x = 1 $.

Therefore, the Ratio Test is inconclusive for any value of $ x $. To find the interval of convergence, we need to consider the behavior of the series at its endpoints.

When $ x = 1 $, the series becomes the harmonic series, which is known to diverge. Therefore, the interval of convergence starts from $ x > 1 $.

When $ x > 1 $, the series converges by the p-series test. Therefore, the interval of convergence extends to $ x > 1 $.

Thus, the interval of convergence for the series $ \sum_{n=1}^{\infty} \frac{3}{n^x} $ is $ x > 1 $.