Series converges or diverges

Question

∞

∑ (2n+3)/(n²+3n+1)

n=0

Answer 1

While ignoring lower order polynomial terms works to give a quick yes/no result, the method may not get full credit if you are asked to show your work using the comparison test.   Ignoring lower order term may produce a series that may be larger than the series to which you want to make your comparison.   In the case where you are trying to prove divergence, comparing your result to a larger series is not conclusive.     You need to producing a smaller, diverging series for this problem.   Dropping the 3 from 2n+3 works, but dropping the lower order 3n+1 from the denominator makes the denominator smaller and the overall expression larger.   Not good.   A different strategy is needed for the denominator.

 

One method that is acceptable is to use the comparison test.

 

we can note that   2n < 2n+3 for all n

and that  the denominator (n²+3n + 1) is always less than 4n² for any n >  2

 

Decreasing the numerator while increasing the denominator is guaranteed to produce a smaller fraction.  Accordingly we can compare say that each of the terms of the given series are greater than  2n/4n²  or 1/2n for all n > 2.    Since 1/n diverges, then (1/2)*(1/n) also diverges, and our original series is greater than a known divergent series.   This method of applying the comparison test is known to be sufficient for a calculus test.   

Answer 2

Ignore all terms except 2n in the numerator and n² in the denominator.

 

Let a_n = (2n+3)/(n²+3n+1)    b_n = 2n/n² = 2/n

 

a_n and b_n are positive for all n > 0

 

Lim_n→∞ (a_n/b_n)  = lim_n→∞ [(2n²+3n)/(2n²+3n+1)] = 1

 

Since the limit is finite and positive, then, by the Limit Comparison Test, ∑a_n (n=1 to ∞) and ∑b_n (n = 1 to ∞) both converge or both diverge.  ∑b_n (n=1 to ∞) diverges because it is a multiple of the harmonic series.  Therefore, ∑a_n (n = 1 to ∞) also diverges.

 

So, ∑a_n (n=0 to ∞) = a₀ + ∑a_n (n=1 to ∞) diverges.