MHB Determine the convergence of the series

[Nikolas7]

Need help. Determine the convergence of the series:
1. sum (Sigma E) from n=1 to infinity of: 1/((2*n+3)*(ln(n+9))^2))
2. sum (Sigma E) from n=1 to infinity of: arccos(1/(n^2+3))
I think the d'alembert is unlikely to help here.

 

[Prove It]

Need help. Determine the convergence of the series:
1. sum (Sigma E) from n=1 to infinity of: 1/((2*n+3)*(ln(n+9))^2))
2. sum (Sigma E) from n=1 to infinity of: arccos(1/(n^2+3))
I think the d'alembert is unlikely to help here.

1. is apparently convergent by comparison. Trying to think of a function to compare it to.

2. By definition of the inverse cosine function, we have $\displaystyle \begin{align*} 0 \leq \arccos{ \left( x \right) } \leq \pi \end{align*}$. Can you use this to create a comparison?

 

[Nikolas7]

Thanks for your reply.
1. What is function for compare? I don't know yet. I confused by the square of ln.
2. I try to change arrcos on arcsin. Can i use arcsinx=pi/2-arccosx?

 

[Opalg]

Thanks for your reply.
1. What is function for compare? I don't know yet. I confused by the square of ln.

The general principle here is that $\ln n$ goes to infinity slower than any positive power of $n$. So for example $\ln n < n^{1/ 4}$ for all sufficiently large $n$. Therefore $(\ln n)^2 < n^{1/ 2}$, so that $\dfrac1{n(\ln n)^2} > \dfrac1{n^{3/2}}$.

2. I try to change arrcos on arcsin. Can i use arcsinx=pi/2-arccosx?

You could look at it that way. If $n$ is large then $\dfrac1{n^2+3}$ is close to $0$ and therefore so is $\arcsin\Bigl( \dfrac1{n^2+3} \Bigr).$ What does that tell you about $\arccos\Bigl( \dfrac1{n^2+3} \Bigr)$?