Question

what is the Direct Comparison Test

Answer 1

I'm assuming you're studying series in Calculus 2?

Answer 2

yes and I'm failing at this module miseralbly

Answer 3

Yea series can be very tricky sometimes. Anyway, the Direct Comparison Test is:\[\sum_{n=1}^{\infty} b_{n}\] which converges, implies that: \[\sum_{n=1}^{\infty} a_{n} \] also converges if: \[a_{n}\le b _{n}, \forall n \in \mathbb{N} \]This also holds true to test for divergence. Have you had much practice with applying these?

Answer 4

I'll demonstrate how this works below to try and get you to see how this is applied to a problem:
Let's say we wish to determine if the series below converges:
\[\sum_{n=1}^{\infty}\frac{ 1 }{ n+3^n }\]We would first compare this to a series which is known. In this case, ignore the n and just focus on the 1/3^n. That looks just like a p-series, so let's compare this one to a similar p-series.
\[\sum_{n=1}^{\infty}\frac{ 1 }{ n+3^n } \rightarrow \sum_{n=1}^{\infty}\frac{ 1 }{ 3^n }\]Now we know, or it would be wise to remember, that a p-series ALWAYS converges so long as n>1, otherwise it's a divergent harmonic series.

Since we have determined that 1/3^n converges, we apply the DCT to see if the one we are inquiring about converges as well. This means the condition An <= Bn must be satisfied.
\[\frac{ 1 }{ n+3^n } \le \frac{ 1 }{ 3^n }\]By studying this, we can notice that for any value of n, the fraction on the left will be smaller, because the denominator will always be larger. In this way, we have shown that An <= Bn for all values of n. That concludes the DCT and proves that:
\[\sum_{n=1}^{\infty}\frac{ 1 }{ n+3^n }\] is convergent!