How do you determine if #Sigma (2n-3)/(5n+6)# from #n=[0,oo)# converge and find the sums when they exist?

Question

Elijah Abram · Accepted Answer

The sum diverges.

Sum of an infintite series #sum_(n=0)^ooa_n# converges only if #lim_(n->oo)a_n=0#. In this question, #lim_(n->oo)(2n-3)/(5n-6)=2/5≠0# and the sum diverges to infinity.

Paisley Johnson · Answer

undefined To determine convergence of the series Σ(2n-3)/(5n+6) from n=0 to infinity, you can use the ratio test.

1. Apply the ratio test: 
   Limit as n approaches infinity of |(a_(n+1))/(a_n)|.
   Where a_n = (2n-3)/(5n+6).

2. Find a_(n+1): 
   a_(n+1) = (2(n+1)-3)/(5(n+1)+6).

3. Substitute a_n and a_(n+1) into the ratio test:
   Limit as n approaches infinity of |(2(n+1)-3)/(5(n+1)+6) / ((2n-3)/(5n+6))|.

4. Simplify the expression:
   Limit as n approaches infinity of |(2n+2-3)/(5n+11) * (5n+6)/(2n-3)|.
   = Limit as n approaches infinity of |(2n-1)(5n+6)/(5n+11)(2n-3)|.

5. Take the limit:
   Using the properties of limits, the limit evaluates to 10/10 = 1.

6. Analyze the result:
   Since the limit is equal to 1, the ratio test is inconclusive.

7. Apply another convergence test:
   You can try using the alternating series test if appropriate, or another convergence test such as the integral test.

For finding the sum when it exists, if the series converges, you can evaluate the sum using appropriate methods such as partial fraction decomposition, geometric series sum formula, or other techniques depending on the nature of the series.