Determine the convergence of the series

In summary, we discussed the convergence of two series - one involving the function $1/((2*n+3)*(ln(n+9))^2)$ and the other involving the inverse cosine function. We determined that the first series is convergent by comparison, and for the second series, we utilized the definition of the inverse cosine function and determined that it is also convergent.
  • #1
Nikolas7
22
0
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.
 
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  • #2
Nikolas7 said:
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?
 
  • #3
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?
 
  • #4
Nikolas7 said:
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 $\dfrac1{n(\ln n)^2} > \dfrac1{n^{3/2}}$.

Nikolas7 said:
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 $\arccos\Bigl( \dfrac1{n^2+3} \Bigr)$?
 

Related to Determine the convergence of the series

What is the meaning of convergence in a series?

Convergence in a series refers to the behavior of the series as the number of terms increases. A series is said to converge if the terms approach a finite limit as the number of terms increases, and diverge if the terms do not approach a limit.

How do you determine the convergence of a series?

The convergence of a series is determined by analyzing the behavior of its terms. This can be done through various tests, such as the ratio test, comparison test, or the integral test. These tests help determine if the series converges, diverges, or is inconclusive.

What is the difference between absolute and conditional convergence?

Absolute convergence refers to a series where the terms are always positive and the series converges. Conditional convergence refers to a series where the terms alternate in sign and the series converges. In both cases, the series converges, but absolute convergence is considered stronger as it is not dependent on the order of the terms.

What happens when a series diverges?

When a series diverges, it means that the terms do not approach a finite limit as the number of terms increases. This could be due to the terms increasing without bound, alternating between positive and negative values, or not following a specific pattern. In such cases, the series is said to be divergent and does not have a defined sum.

Why is it important to determine the convergence of a series?

Determining the convergence of a series is important in many areas of mathematics and science. It helps in evaluating infinite sums, solving differential equations, and analyzing the behavior of functions. It also allows for the identification of patterns and relationships between terms, leading to a better understanding of the underlying concepts.

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