# Power Series Convergence Assistance

• MHB
• theCalc
In summary, the given power series converges to the number 1, as shown by manipulating the series and using the fact that it can be written in terms of the Maclaurin series for e^-1. The Ratio Test also confirms the convergence of the series.
theCalc
The power series

$$\sum_{n = 2}^\infty \frac{(n-1)(-1)^n}{n!}$$

converges to what number?

So far, I've tried using the Ratio Test and the limit as n approaches infinity equals $0$. Also since $L<1$, the power series converges by the Ratio Test.

Last edited by a moderator:
Maclaurin tells us:

$$\displaystyle e^{-1}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}$$

You can write the given series in terms of the above. :)

I'm a bit lost and confused as to how I would manipulate the series, could you explain what steps would I need to take? Thanks.

Well, I think the first thing I would do is write:

$$\displaystyle e^{-1}=1-1+\sum_{n=2}^{\infty}\frac{(-1)^n}{n!}=\sum_{n=2}^{\infty}\frac{(-1)^n}{n!}$$

Next, let's write:

$$\displaystyle S=\sum_{n=2}^{\infty}\left(\frac{(-1)^n(n-1)}{n!}\right)=\sum_{n=2}^{\infty}\left(\frac{(-1)^nn}{n!}\right)-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)$$

$$\displaystyle S=\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{(n-1)!}\right)-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)$$

$$\displaystyle S=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n!}\right)-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)$$

$$\displaystyle S=1-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)$$

$$\displaystyle S=1-2\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)=\,?$$

## 1. What is a power series?

A power series is an infinite series of the form ∑n=0∞ cn(x-a)n, where cn and a are constants and x is a variable. It is a type of mathematical series commonly used in calculus and mathematical analysis.

## 2. How do I determine if a power series converges?

The convergence of a power series depends on the value of x. One way to determine convergence is to use the ratio test, which states that if limn→∞|cn+1(x-a)n+1/cn(x-a)n| < 1, then the series converges. However, this test is not always conclusive and other methods, such as the root test, may need to be used.

## 3. What is the interval of convergence for a power series?

The interval of convergence is the range of values for x that allows a power series to converge. It can be found by using convergence tests, such as the ratio or root test, and determining the values of x that satisfy the conditions for convergence. The interval may include the center point of the series or it may be a range of values around the center point.

## 4. Can a power series diverge?

Yes, a power series can diverge if the value of x does not fall within the interval of convergence. In this case, the series will not have a finite value and will either approach infinity or alternate between positive and negative infinite values.

## 5. How can I use power series to approximate functions?

Power series can be used to approximate a wide range of functions, including trigonometric, logarithmic, and exponential functions. By manipulating the coefficients and the center point of the series, it is possible to find a power series that closely matches the given function within a certain range of values for x. This can be useful in solving mathematical problems and making approximations in scientific calculations.

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