Question

4 (2) This question is about the series n2 + 4n +3 n=1 (a) Show that this series converges, using the integral test. (Hint: Partial fraction decomposition.) (b) Notice this is not a geometric series, so we shouldn't expect to know what it converges to. But use the decomposition 4 into the difference n2 4n of two sums. (c) Use index shifts to make these sums looks similar enough to rewrite this expression without Σ. 4 (d) Take the limit as B+ 0 to find n2 + 4n +3 B from part (a) to break m2 + An + 3 n=1 n=1 (2) 10

Answer 1

(a) Given: f(x) = x^2 + 4x + 3.

The partial fraction decomposition of f(x) is:

f(x) = (x+1)(x+3)

Now, we need to find the integral of this function from 1 to infinity:

∫[1,∞] (x+1)(x+3) dx

Since the integral converges, we can conclude that the series also converges.

(b) This series is not geometric, so we don't know what it converges to. However, we can decompose the given series as the difference of two sums:

Σ(n^2 + 4n + 3) = Σ(n^2) - Σ(4n)

(c) We can use index shifts to make these sums look similar enough to rewrite the expression without Σ:

Σ(n^2) - Σ(4n) = Σ(n^2 - 4n)

(d) To find the limit as B approaches 0, we can evaluate the limit of the expression n^2 + 4n + 3:

lim(B→0) (n^2 + 4n + 3) = n^2 + 4n + 3

So, the limit of the series is n^2 + 4n + 3.