How do you determine whether #sum n/3^(n+1)# from 1 to infinity converges or diverges?

Question

Aurora Alvarado · Accepted Answer

The Ratio Test will show it converges. The Ratio Test says: 
The series #sum_{n=1}^{\infty} a_n# converges if  #\lim_{n 	o \infty} \abs\frac{a_{n+1}}{a_n} < 1#, diverges if  #\lim_{n 	o \infty} \abs\frac{a_{n+1}}{a_n} > 1#, and is inconclusive if #\lim_{n 	o \infty} \abs\frac{a_{n+1}}{a_n} = 1#. For your specific problem,  #a_n = \frac{n}{3^{n+1}}# and #a_{n+1} = \frac{n+1}{3^{n+2}}# so #\lim_{n 	o \infty} \abs\frac{a_{n+1}}{a_n} = \lim_{n 	o \infty} \abs\frac{\frac{n+1}{3^{n+2}}}{\frac{n}{3^{n+1}}} = #
# =  \lim_{n 	o \infty} \abs{\frac{n+1}{3^{n+2}} \cdot \frac{3^{n+1}}{n}} = \lim_{n 	o \infty} \abs{\frac{n+1}{3n} } =#
# = \frac{1}{3}\lim_{n 	o \infty} \abs{\frac{n+1}{n} }  = \frac{1}{3} < 1#, so the series converges.

Axel Alvarez · Answer

undefined To determine whether the series $ \sum \frac{n}{3^{n+1}} $ from $ n = 1 $ to $ \infty $ converges or diverges, you can use the ratio test.

The ratio test states that if $ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1 $, then the series converges; if $ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} > 1 $ or the limit is undefined, then the series diverges; and if $ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 1 $, the test is inconclusive.

For the given series:
$$ a_n = \frac{n}{3^{n+1}} $$

To apply the ratio test, we compute:
$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+1)}{3^{n+2}} \cdot \frac{3^{n+1}}{n} = \lim_{n \to \infty} \frac{n+1}{3n} $$

Simplify and compute the limit:
$$ \lim_{n \to \infty} \frac{n+1}{3n} = \frac{1}{3} < 1 $$

Since the limit is less than 1, by the ratio test, the series $ \sum \frac{n}{3^{n+1}} $ converges.