Direct Comparison Test

Question

Use the Direct Comparison Test to determine if each series converges or diverges.

∑ (Infinity)(n=1) (√(n+4)/(n^4+4))

Please step by step explain, as I am so confused!

Answer 1

We'll compare the given series to √(n/n^4) or √(1/n^3)=1/n^3/2. Which we know converges since p=3/2>1. Thus we want some upper bound C/(n^3/2) on √((n+4)/(n^4+4)). ie √((n+4)/(n^4+4))≤(C/(n^3/2)) for large n. Atp we want to turn (n+4)/(n^4+4) into some d/(n^3/2) so that when we take the square root it has the form (√d)/(n^3/2). ie our upper bound. To do this √((n+4)/(n^4+4))≤sqrt(2n/(n^4)) which holds for n≥4. √(2n/(n^4))=(√2)/(n^3/2). We know this converges bc p=3/2>1. Since 0<√((n+4)/(n^4+4))≤(√2)/(n^3/2) for n≥4 and ∑ (Infinity)(n=1) ((√2)/(n^3/2)) then we know the original series converges by direct comparison test.

Note: if your teacher is super strict she might say to include the first few terms of the series ie n=1 to n=3 since for these values our series is larger than the comparison. But this likely suffices.