Series Convergence with Comparison Test

[Zoey93]

Hey,

I am working on Calculus III and Analysis, I really need help with this one problem. I am not even sure where to begin with this problem. I have attached my assignment to this thread and the problem I need help with is A. Thank you!

 

[Opalg]

Hey,

I am working on Calculus III and Analysis, I really need help with this one problem. I am not even sure where to begin with this problem. I have attached my assignment to this thread and the problem I need help with is A. Thank you!

Here are a couple of hints to get you started.

1. The convergence or divergence of a series is not affected by the first few terms. So you can neglect the first three terms of the series, and concentrate on those starting with the term $\frac1{3+\sqrt2}.$

2. From that point onwards, the denominator of each term consists of two elements: a power of 3 and a square root. One of those elements is going to be much larger than the other. So think about what the series would look like if you neglected the smaller element.

 

[Prove It]

Each of the important terms are of the form $\displaystyle \begin{align*} \frac{1}{ 3^n + \sqrt{n+1}} \end{align*}$, and since $\displaystyle \begin{align*} 3^n + \sqrt{n + 1} > 3^n \end{align*}$ for $\displaystyle \begin{align*} n \geq 0 \end{align*}$, that means $\displaystyle \begin{align*} \frac{1}{3^n + \sqrt{n + 1}} < \frac{1}{3^n} = \left( \frac{1}{3} \right) ^n \end{align*}$. So you can make a comparison to a simple geometric series :)