# Calculating where X converges

• DMOC
In summary, the question is asking for values of x that will cause the series \sum \frac{(2^k)(x^k)}{ln(k + 2)} to converge as k goes from 0 to infinity. The homework equations used to solve this problem were the Limit Test and the Alternating Series Test. The attempt at a solution involved using the Limit Comparison Test, which led to the conclusion that both endpoints of the series, x=1/2 and x=-1/2, converge. However, the answer key states that x=1/2 diverges while x=-1/2 converges. It was later realized that the Alternating Series Test cannot be used with a non-alternating series, and the

## Homework Statement

Determine values for x where $$\sum \frac{(2^k)(x^k)}{ln(k + 2)}$$ converges if k goes from 0 to infinity.

## Homework Equations

Limit test and alternating series test.

## The Attempt at a Solution

I used the limit comparison test to test out ak+1 and an and I ended up getting $$(-1/2) \le x \le (1/2)$$

Then I had to compare the endpoints. I noticed that the numerator is always 1 in the case of x=1/2 or always 1 or -1 (in the case of x = -1/2).

So by the alternating series test, I concluded that BOTH of the endpoints converges. However, my answer key says that x=1/2 diverges while x=-1/2 converges (I'm correct there).

FYI, for x = 1/2,

I said that an+1 < an and that the limit as k -> $$\infty$$ = 0, so why does it apparently diverge?

I'm with you at x=(-1/2), but how did you conclude 1/ln(k+2) converges by the alternating series test? It doesn't even alternate.

Wait so I can't use the alternating series test with a non alternating series ... wow I'm going to kick myself in the @%$@# for overlooking this. So I assume I just use like a comparison test for 1/ln(k+2) with 1/k (diverges) since ln makes values of k smaller so the frac. gets bigger. DMOC said: Wait so I can't use the alternating series test with a non alternating series ... wow I'm going to kick myself in the @%$@# for overlooking this.

So I assume I just use like a comparison test for 1/ln(k+2) with 1/k (diverges) since ln makes values of k smaller so the frac. gets bigger.

Right.

DMOC said:
I used the limit comparison test to test out ak+1 and an and I ended up getting $$(-1/2) \le x \le (1/2)$$

This is the Ratio Test not the Limit Comparison Test.
Also, you should write $$a_k$$ not $$a_n$$.

## What is the concept of convergence in mathematics?

The concept of convergence in mathematics refers to the behavior of a sequence of numbers or functions as it approaches a specific value or point. It is a fundamental concept in calculus and is used to determine the behavior of various mathematical operations.

## How is the convergence of a sequence or series determined?

The convergence of a sequence or series is typically determined by analyzing the behavior of its terms as the number of terms increases. This can involve finding a limit, checking for the existence of a pattern, or using various convergence tests such as the ratio test or comparison test.

## What is the importance of determining the convergence of a sequence or series?

Determining the convergence of a sequence or series is important in mathematics because it allows us to understand the behavior of a mathematical operation and make predictions about its outcome. It also helps us to ensure the accuracy and validity of mathematical calculations and proofs.

## What are some common examples of convergence in mathematics?

Some common examples of convergence in mathematics include the convergence of geometric series, power series, and Taylor series. Other examples include the convergence of sequences and the convergence of integrals.

## What are some common methods for calculating where a sequence or series converges?

Some common methods for calculating where a sequence or series converges include using convergence tests such as the ratio test or comparison test, finding a limit of the sequence or series, and checking for the existence of a pattern in the terms of the sequence or series.