# Convergence of Sum 1/n(n+1) * (sin(x))^n

• stukbv
In summary, the given sum converges for all x in the reals and can be written as sum 1/n(n+1) * (sin(x))^n. It can be shown to be differentiable on a certain interval by letting x = sinx and differentiating to get 1/(n+1) * x^n-1. However, the correct derivative should be calculated for the given sum, which is \sum_{n=0}^{+\infty}{\frac{1}{n(n+1)}\sin^n(x)}.
stukbv

## Homework Statement

Sum 1/n(n+1) * (sin(x))^n .
Show this converges for all x in the reals.
Find with proof an interval on which it determines a differentiable function of x together with an expression of its derivative in terms of standard functions.

## The Attempt at a Solution

I have showen convergence via a comparison to sum 1/n(n+1) so this is okay,
I think now we let x=sinx and write it like ;
Sum 1/n(n+1) * x^n .
Differentiating this gives you 1/(n+1) * x^n-1 now basically am stuck again :(

Thankyou!

Hi stukbv!

stukbv said:
I have showen convergence via a comparison to sum 1/n(n+1) so this is okay,

Sounds good!

I think now we let x=sinx and write it like ;
Sum 1/n(n+1) * x^n .
Differentiating this gives you 1/(n+1) * x^n-1 now basically am stuck again :(

No, now you're calculating a different derivative. You really need to calculate the derivative of

$$\sum_{n=0}^{+\infty}{\frac{1}{n(n+1)}\sin^n(x)}$$

## 1. What is the formula for the convergence of Sum 1/n(n+1) * (sin(x))^n?

The formula for the convergence of Sum 1/n(n+1) * (sin(x))^n is given by the ratio test, where we evaluate the limit as n approaches infinity of the absolute value of (an+1/an), with an = 1/n(n+1) * (sin(x))^n. If the limit is less than 1, the series converges, and if the limit is greater than 1, the series diverges.

## 2. How do you determine the interval of convergence for Sum 1/n(n+1) * (sin(x))^n?

The interval of convergence for Sum 1/n(n+1) * (sin(x))^n can be determined by using the ratio test. We first find the interval of convergence for the infinite series of (sin(x))^n, which is -1 < x < 1. Then, we use the ratio test to find the values of x that make the limit less than 1, which gives us the interval of convergence for the entire series.

## 3. Can the series Sum 1/n(n+1) * (sin(x))^n converge for all values of x?

No, the series Sum 1/n(n+1) * (sin(x))^n cannot converge for all values of x. As mentioned in the previous question, the interval of convergence for the infinite series of (sin(x))^n is -1 < x < 1. Therefore, for any values of x outside of this interval, the series will diverge.

## 4. What is the significance of the alternating signs in the series Sum 1/n(n+1) * (sin(x))^n?

The alternating signs in the series Sum 1/n(n+1) * (sin(x))^n indicate that it is an alternating series, meaning that the signs of the terms alternate between positive and negative. This is important because it allows us to use the alternating series test to determine the convergence of the series. If the terms of an alternating series decrease in absolute value and approach 0, then the series will converge.

## 5. How can the series Sum 1/n(n+1) * (sin(x))^n be used in real-life applications?

The series Sum 1/n(n+1) * (sin(x))^n can be used in real-life applications to model oscillating phenomena, such as sound or electrical signals. The alternating signs and the sinusoidal term make it a good fit for representing these types of periodic functions. It can also be used in the study of differential equations and in analytical calculations involving infinite series.

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