How do you show that the series #1/4+1/7+1/10+...+1/(3n+1)+...# diverges?

Question

Charles Anctil · Accepted Answer

The series:

#sum_(n=1)^oo 1/(3n+1)#

is divergent.

Let #s_n# be the #n#-th partial sum of the series, and consider the sums with index #n=3^k-1#: #sigma_k = s_(3^n-1) = sum_(i=1)^(3^n-1) 1/(3i+1)# Now we evaluate the difference between two consecutive such partial sums: #sigma_(k+1) = sigma_k + 1/(3(3^n) +1) + 1/(3(3^n+1) +1) + ... + 1/(3(3^(n+1)-1) +1)# We can see that all the terms that were added are larger than #1/3^(n+2)# and their number is: #3^(n+1) - 1 -3^n +1 = 2*3^n#, so we have: #sigma_(k+1) >= sigma_k + (2*3^n)*1 /3^(n+2)# or: #sigma_(k+1) >= sigma_k + 2/9# Now if we start from: # sigma_1 = s_2 = 1/4+1/7 = 11/28 > 2/9# then we now that: # sigma_2 >= sigma_1 +2/9 >2*2/9# # sigma_3 >= sigma_2 +2/9 >3*2/9# and in general: # sigma_k > k*2/9# so that we have: #lim_(k->oo) sigma_k = oo# and since the #sigma_k# are a subset of the partial sums of the series, then the series is proven to be divergent.

Christopher Allers · Answer

undefined To show that the series $ \frac{1}{4} + \frac{1}{7} + \frac{1}{10} + \ldots + \frac{1}{3n+1} + \ldots $ diverges, we can use the limit comparison test or the integral test.

Using the Limit Comparison Test:
1. Choose a series $ b_n $ that is known to diverge.
2. Take the limit as $ n $ approaches infinity of the ratio $ \frac{a_n}{b_n} $, where $ a_n $ is the given series.
3. If the limit is greater than zero or infinite, then both series either converge or diverge together.

Using the Integral Test:
1. Consider the function $ f(x) = \frac{1}{3x + 1} $.
2. Check if the function is continuous, positive, and decreasing for all $ x \geq 1 $.
3. Integrate the function from 1 to infinity. If the integral diverges, then the series also diverges.

For this specific series, let's use the Limit Comparison Test with the harmonic series $ \sum \frac{1}{n} $, which is known to diverge.

$$ \lim_{n \to \infty} \frac{\frac{1}{3n+1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{3n+1} = \frac{1}{3} $$

Since $ \frac{1}{3} $ is a positive finite number, and the harmonic series diverges, the given series also diverges by the Limit Comparison Test.